Sequences, Series, and Combinatorics

Sequences and Factorial Notation

In mathematics, a sequence is an ordered list of numbers. Each number in the list is called a term. Because order matters, the position of each term is important.

For example, in the sequence \(2, 4, 6, 8, 10, \dots\), the first term is \(2\), the second term is \(4\), and the third term is \(6\).

Sequences can be described in two main ways:

• with an explicit formula, which gives the value of the \(n\)th term directly,
• with a recursive formula, which tells how to get each term from the one before it.

This lesson also introduces factorial notation, written with an exclamation mark. Factorials appear often in sequences, counting problems, and combinatorics.

1. Understanding sequences

A sequence is usually written as \(a_1, a_2, a_3, \dots\), where \(a_n\) means the term in position \(n\).

For example, if a sequence begins \(5, 8, 11, 14, \dots\), then:

• \(a_1 = 5\)
• \(a_2 = 8\)
• \(a_3 = 11\)
• \(a_4 = 14\)

This sequence increases by \(3\) each time.

2. Explicit formulas

An explicit formula gives the value of a term directly from its position number \(n\). This means you do not need to know the term before it.

For the sequence \(5, 8, 11, 14, \dots\), an explicit formula is

This works because the sequence starts at \(5\) and adds \(3\) for each new term.

We can simplify this formula:

So another explicit formula is

If you want the 10th term, substitute \(n=10\):

Explicit formulas are useful when you want to jump straight to a particular term.

3. Recursive formulas

A recursive formula defines a sequence using one or more earlier terms. To use a recursive rule, you must know the starting term.

For the sequence \(5, 8, 11, 14, \dots\), a recursive definition is:

This says:

• start with \(5\),
• then add \(3\) each time to get the next term.

Recursive formulas are useful when a pattern is built step by step.

4. Explicit vs recursive definitions

Both forms describe the same sequence, but they do it differently.

• Explicit: good for finding any term quickly.
• Recursive: good for showing how the pattern grows from one term to the next.

For example, the sequence \(2, 4, 8, 16, 32, \dots\) can be written as:

Explicit form:

This works because:

• \(a_1 = 2^1 = 2\)
• \(a_2 = 2^2 = 4\)
• \(a_3 = 2^3 = 8\)

Recursive form:

This says each term is twice the previous term.

5. Recognizing common sequence patterns

Two common types of sequences in Grade 12 mathematics are:

• Arithmetic sequences: add the same number each time.
• Geometric sequences: multiply by the same number each time.

Arithmetic example: \(7, 10, 13, 16, \dots\)

• common difference: \(3\)
• recursive form: \(a_1=7\), \(a_n=a_{n-1}+3\)
• explicit form: \(a_n=7+3(n-1)\)

Geometric example: \(3, 6, 12, 24, \dots\)

• common ratio: \(2\)
• recursive form: \(a_1=3\), \(a_n=2a_{n-1}\)
• explicit form: \(a_n=3(2^{n-1})\)

6. Factorial notation

A factorial is the product of all positive integers from \(1\) up to a given whole number.

It is written using an exclamation mark.

For a positive integer \(n\):

Examples:

• \(1! = 1\)
• \(2! = 2 \cdot 1 = 2\)
• \(3! = 3 \cdot 2 \cdot 1 = 6\)
• \(4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24\)
• \(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\)

A special fact to remember is:

This may seem unusual, but it is a standard definition in mathematics and is very useful in combinatorics and formulas.

7. Factorials as recursive definitions

Factorials can also be defined recursively. This connects factorial notation directly to recursive sequences.

This means each factorial is found by multiplying the number by the factorial before it.

For example:

This is a recursive pattern because each new value depends on the previous one.

8. Sequences involving factorials

Some sequences use factorial notation in their formulas. For example:

This gives the sequence:

because:

• \(a_1 = 1! = 1\)
• \(a_2 = 2! = 2\)
• \(a_3 = 3! = 6\)
• \(a_4 = 4! = 24\)
• \(a_5 = 5! = 120\)

This sequence grows very quickly.

9. Worked Examples

Example 1: Write an explicit and recursive formula

Consider the sequence \(4, 9, 14, 19, \dots\).

Step 1: Find the pattern.

Each term increases by \(5\).

Explicit formula:

Simplify:

So the explicit formula is

Recursive formula:

Example 2: Use an explicit formula to find a term

Suppose a sequence is defined by

Find \(a_5\).

Substitute \(n=5\):

So the 5th term is \(112\).

Example 3: Generate terms from a recursive formula

A sequence is defined by

Find the first five terms.

Start with the first term:

• \(a_1 = 3\)
• \(a_2 = a_1 + 4 = 3 + 4 = 7\)
• \(a_3 = a_2 + 4 = 7 + 4 = 11\)
• \(a_4 = a_3 + 4 = 11 + 4 = 15\)
• \(a_5 = a_4 + 4 = 15 + 4 = 19\)

So the first five terms are:

Example 4: Evaluate factorials and connect them to a sequence

Let \(a_n = (n+1)!\). Find \(a_1\), \(a_2\), \(a_3\), and \(a_4\).

Substitute each value of \(n\):

• \(a_1 = (1+1)! = 2! = 2\)
• \(a_2 = (2+1)! = 3! = 6\)
• \(a_3 = (3+1)! = 4! = 24\)
• \(a_4 = (4+1)! = 5! = 120\)

So the sequence begins:

10. Common mistakes to avoid

• Confusing term number with term value. For example, \(a_3\) means the 3rd term, not necessarily the number 3.
• Forgetting the starting term in a recursive formula. A recursive rule is incomplete without an initial value like \(a_1\).
• Using factorial incorrectly. Remember that \(5!\) means \(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\), not \(5\).
• Forgetting that \(0! = 1\).
• Mixing up arithmetic and geometric patterns. Arithmetic sequences add; geometric sequences multiply.

11. How to decide which formula to use

When you are given a sequence, ask these questions:

1. Does the pattern add the same amount each time?
2. Does the pattern multiply by the same amount each time?
3. Do I need a direct formula for the \(n\)th term, or a step-by-step rule?

If you want to find a faraway term quickly, an explicit formula is usually best.

If the sequence is naturally built from previous terms, a recursive formula may be more useful.

Brief Summary

A sequence is an ordered list of numbers, and each term can be described by its position. An explicit formula gives the \(n\)th term directly, while a recursive formula defines each term using earlier terms and a starting value.

Factorial notation, written as \(n!\), means multiplying all positive integers from \(n\) down to \(1\). Factorials can also be written recursively, which connects them closely to recursive sequences. Understanding both types of sequence definitions helps you describe patterns clearly and solve problems in sequences, series, and combinatorics.

Arithmetic Sequences and Series are important tools in algebra because they describe patterns that change by the same amount each time. You may see them in savings plans, seating arrangements, step patterns, and many other real-life situations.

In this lesson, you will learn how to recognize an arithmetic sequence, write a rule for its terms, find any term in the sequence, and calculate the sum of a finite arithmetic series.

A sequence is an ordered list of numbers. Each number in the list is called a term.

For example, in the sequence \(5, 8, 11, 14, 17\), the terms increase by \(3\) each time.

A sequence is called arithmetic if the difference between consecutive terms is always the same. This fixed amount is called the common difference.

In the example \(5, 8, 11, 14, 17\), the common difference is

Since the difference is constant, this is an arithmetic sequence.

How to identify an arithmetic sequence:

• Subtract one term from the next.
• If the difference is always the same, the sequence is arithmetic.
• If the differences change, the sequence is not arithmetic.

Examples:

• \(2, 6, 10, 14, 18\) is arithmetic because the difference is \(+4\).
• \(20, 15, 10, 5, 0\) is arithmetic because the difference is \(-5\).
• \(3, 6, 12, 24\) is not arithmetic because the differences are not constant.

To describe an arithmetic sequence, we usually use:

• \(a_1\) for the first term
• \(d\) for the common difference
• \(a_n\) for the \(n\)th term

The formula for the \(n\)th term of an arithmetic sequence is

This formula works because you begin at the first term and add the common difference \((n-1)\) times.

For example, if \(a_1=4\) and \(d=3\), then

So the sequence is \(4, 7, 10, 13, 16, \dots\)

Recursive form is another way to write an arithmetic sequence. It tells you how to get each term from the one before it.

For an arithmetic sequence, the recursive rule is:

For the sequence \(4, 7, 10, 13, \dots\), the recursive form is

This means start at \(4\), then keep adding \(3\).

Arithmetic series is different from an arithmetic sequence. A sequence is a list of terms, while a series is the sum of those terms.

For example:

• Sequence: \(4, 7, 10, 13\)
• Series: \(4+7+10+13\)

To find the sum of the first \(n\) terms of an arithmetic sequence, use the arithmetic series formula:

This formula says that the sum equals half the number of terms times the sum of the first and last terms.

You can also use an equivalent form when the last term is not known yet:

Both formulas give the same answer. Choose the one that matches the information you know.

Why the sum formula works:

Suppose the series is

If you pair the first and last terms, then the second and second-last terms, each pair has the same total:

This repeating structure is why arithmetic series can be added quickly with a formula instead of adding term by term.

Worked Example 1: Find the common difference and next terms

Determine whether \(12, 17, 22, 27, \dots\) is arithmetic, and find the next two terms.

Step 1: Check the differences.

The difference is always \(5\), so the sequence is arithmetic.

Step 2: Add \(5\) to continue the pattern.

Answer: The sequence is arithmetic with common difference \(d=5\), and the next two terms are \(32\) and \(37\).

Worked Example 2: Find the \(n\)th term

Find the 20th term of the arithmetic sequence \(3, 7, 11, 15, \dots\)

Step 1: Identify the first term and common difference.

Step 2: Use the formula \(a_n=a_1+(n-1)d\).

Answer: The 20th term is \(79\).

Worked Example 3: Write explicit and recursive rules

Write both the explicit formula and recursive formula for the sequence \(18, 13, 8, 3, -2, \dots\)

Step 1: Identify the first term and common difference.

Step 2: Write the explicit formula.

Step 3: Write the recursive formula.

Answer:

• Explicit formula: \(a_n = 18 - 5(n-1)\)
• Recursive formula: \(a_1=18,\; a_n=a_{n-1}-5\)

Worked Example 4: Find the sum of a finite arithmetic series

Find the sum of the first 25 terms of the arithmetic sequence \(6, 10, 14, 18, \dots\)

Step 1: Identify known values.

Step 2: Find the 25th term.

Step 3: Use the series formula.

Answer: The sum of the first 25 terms is \(1350\).

Tips for solving arithmetic sequence and series problems:

• Always check whether the difference between terms is constant.
• Be careful with negative common differences.
• Use \(a_n=a_1+(n-1)d\) to find a specific term.
• Use \(S_n=\frac{n}{2}(a_1+a_n)\) when you know the first and last term.
• Use \(S_n=\frac{n}{2}[2a_1+(n-1)d]\) when you know \(a_1\), \(d\), and \(n\).

Common mistakes to avoid:

• Using \(n\) instead of \(n-1\) in the \(n\)th term formula.
• Confusing a sequence with a series.
• Forgetting that subtracting consecutive terms is how you find the common difference.
• Making sign errors when the sequence decreases.

Quick check:

1. Is \(7, 12, 17, 22, \dots\) arithmetic? What is \(d\)?
2. Find the 15th term of \(9, 13, 17, 21, \dots\)
3. Find the sum of the first 10 terms of \(2, 5, 8, 11, \dots\)

Answers:

1. Yes, it is arithmetic and \(d=5\).
2. \(a_{15}=9+(15-1)(4)=9+56=65\)
3. First find \(a_{10}=2+(10-1)(3)=29\), then $$S_{10}=\frac{10}{2}(2+29)=5\cdot 31=155$$

Summary

An arithmetic sequence has a constant difference between consecutive terms. Its \(n\)th term is found using $$a_n=a_1+(n-1)d$$ and its recursive rule adds the common difference each time.

An arithmetic series is the sum of the terms in an arithmetic sequence. To find the sum of the first \(n\) terms, use $$S_n=\frac{n}{2}(a_1+a_n)$$ or $$S_n=\frac{n}{2}[2a_1+(n-1)d]$$. When you can identify \(a_1\), \(d\), and \(n\), you can solve most arithmetic sequence and series problems accurately.

Geometric Sequences and Series

In many real-life situations, numbers do not increase by adding the same amount each time. Instead, they change by multiplying by the same factor again and again. Patterns like this are called geometric.

For example, the sequence \(3, 6, 12, 24, 48, \dots\) is geometric because each term is found by multiplying the previous term by \(2\).

Understanding geometric sequences and series helps you recognize exponential patterns, write formulas, and calculate totals efficiently. These ideas are also used in finance, population models, science, and technology.

1. What is a geometric sequence?

A geometric sequence is a sequence in which each term is obtained by multiplying the previous term by the same nonzero constant. This constant is called the common ratio.

If the first term is \(a_1\) and the common ratio is \(r\), then the sequence looks like:

$$a_1,\ a_1r,\ a_1r^2,\ a_1r^3,\ \dots$$

For example:

• \(5, 15, 45, 135, \dots\) has common ratio \(r=3\)
• \(80, 40, 20, 10, \dots\) has common ratio \(r=\frac{1}{2}\)
• \(7, -14, 28, -56, \dots\) has common ratio \(r=-2\)

To check whether a sequence is geometric, divide each term by the term before it. If the ratio stays the same, the sequence is geometric.

2. Recursive formula for a geometric sequence

A recursive formula defines each term using the term before it.

For a geometric sequence:

$$a_n = ra_{n-1} \quad \text{for } n \ge 2$$

Along with this, you must know the first term:

$$a_1 = \text{first term}$$

For example, for the sequence \(4, 12, 36, 108, \dots\):

• First term: \(a_1 = 4\)
• Common ratio: \(r = 3\)

Its recursive formula is:

$$a_1=4, \quad a_n = 3a_{n-1}$$

3. Explicit formula for the \(n\)th term

A recursive formula is useful for building the sequence one step at a time, but sometimes you want to find a term directly without listing all previous terms. For that, use the explicit formula.

For a geometric sequence with first term \(a_1\) and common ratio \(r\):

$$a_n = a_1r^{n-1}$$

This works because:

• the first term is \(a_1\)
• the second term is \(a_1r\)
• the third term is \(a_1r^2\)
• the \(n\)th term is \(a_1r^{n-1}\)

4. What is a geometric series?

A sequence is a list of terms. A series is the sum of the terms in a sequence.

So if \(2, 6, 18, 54\) is a geometric sequence, then

$$2+6+18+54$$

is a geometric series.

If we add the first \(n\) terms of a geometric sequence, we get a finite geometric series.

5. Formula for the sum of the first \(n\) terms

If the first term is \(a_1\) and the common ratio is \(r\), then the sum of the first \(n\) terms is:

$$S_n = a_1 + a_1r + a_1r^2 + \dots + a_1r^{n-1}$$

The formula for this sum, when \(r \ne 1\), is:

$$S_n = \frac{a_1(1-r^n)}{1-r}$$

An equivalent version is:

$$S_n = \frac{a_1(r^n-1)}{r-1}$$

Both formulas mean the same thing. You can use whichever one feels easier.

If \(r=1\), then every term is the same, so the sum is simply:

$$S_n = na_1$$

6. Infinite geometric series

Sometimes a geometric pattern continues forever, and we look at the sum of infinitely many terms:

$$a_1 + a_1r + a_1r^2 + a_1r^3 + \dots$$

This is called an infinite geometric series.

Not every infinite geometric series has a finite sum. It only has a finite sum when the terms get closer and closer to \(0\), which happens when:

$$|r|<1$$

In that case, the sum is:

$$S_\infty = \frac{a_1}{1-r}$$

For example, in the series

$$8 + 4 + 2 + 1 + \frac{1}{2} + \dots$$

the first term is \(8\) and the common ratio is \(\frac{1}{2}\), so:

$$S_\infty = \frac{8}{1-\frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16$$

7. Important things to notice about the common ratio

• If \(r>1\), the terms grow larger in size.
• If \(0<r<1\), the terms get smaller but stay positive.
• If \(r<0\), the signs alternate between positive and negative.
• If \(|r|<1\), an infinite geometric series can have a finite sum.

Worked Example 1: Find the common ratio and the next terms

Determine whether the sequence \(2, 10, 50, 250, \dots\) is geometric. If it is, find the common ratio and the next two terms.

Step 1: Divide each term by the previous term.

$$\frac{10}{2}=5, \quad \frac{50}{10}=5, \quad \frac{250}{50}=5$$

The ratio is constant, so the sequence is geometric with common ratio \(r=5\).

Step 2: Find the next terms by multiplying by \(5\).

$$250 \cdot 5 = 1250$$

$$1250 \cdot 5 = 6250$$

Answer: The sequence is geometric, \(r=5\), and the next two terms are \(1250\) and \(6250\).

Worked Example 2: Find the \(n\)th term

Find an explicit formula for the sequence \(81, 27, 9, 3, \dots\), and then find the 7th term.

Step 1: Identify the first term and common ratio.

• \(a_1=81\)
• \(r=\frac{27}{81}=\frac{1}{3}\)

Step 2: Use the explicit formula.

$$a_n = a_1r^{n-1}$$

Substitute the values:

$$a_n = 81\left(\frac{1}{3}\right)^{n-1}$$

Step 3: Find the 7th term.

$$a_7 = 81\left(\frac{1}{3}\right)^6$$

Since \(81=3^4\), we can evaluate directly:

$$a_7 = 81\cdot \frac{1}{729} = \frac{81}{729} = \frac{1}{9}$$

Answer:

$$a_n = 81\left(\frac{1}{3}\right)^{n-1}$$

and the 7th term is

$$a_7 = \frac{1}{9}$$

Worked Example 3: Find the sum of a finite geometric series

Find the sum of the first 6 terms of the geometric sequence \(3, 6, 12, 24, \dots\).

Step 1: Identify the values.

• \(a_1=3\)
• \(r=2\)
• \(n=6\)

Step 2: Use the finite sum formula.

$$S_n = \frac{a_1(1-r^n)}{1-r}$$

$$S_6 = \frac{3(1-2^6)}{1-2}$$

$$S_6 = \frac{3(1-64)}{-1}$$

$$S_6 = \frac{3(-63)}{-1} = 189$$

Check by listing terms:

The first 6 terms are \(3, 6, 12, 24, 48, 96\).

$$3+6+12+24+48+96=189$$

Answer: The sum of the first 6 terms is \(189\).

Worked Example 4: Find the sum of an infinite geometric series

Find the sum of:

$$12 + 6 + 3 + 1.5 + \dots$$

Step 1: Identify the first term and ratio.

• \(a_1=12\)
• \(r=\frac{6}{12}=\frac{1}{2}\)

Step 2: Check whether the infinite sum exists.

Since \(|r|=\frac{1}{2}<1\), the series has a finite sum.

Step 3: Use the infinite sum formula.

$$S_\infty = \frac{a_1}{1-r}$$

$$S_\infty = \frac{12}{1-\frac{1}{2}} = \frac{12}{\frac{1}{2}} = 24$$

Answer: The sum of the infinite geometric series is \(24\).

8. Common mistakes to avoid

• Confusing arithmetic and geometric sequences: Arithmetic sequences add the same amount each time. Geometric sequences multiply by the same ratio each time.
• Using the wrong ratio: Always divide a term by the previous term to find \(r\).
• Forgetting the exponent \(n-1\): In \(a_n=a_1r^{n-1}\), the exponent is one less than the term number.
• Using the infinite sum formula when \(|r|\ge 1\): The formula \(S_\infty=\frac{a_1}{1-r}\) only works when \(|r|<1\).
• Sign errors with negative ratios: If \(r\) is negative, the signs alternate. Be careful when calculating powers.

9. Quick strategy for solving problems

1. Check whether the pattern has a constant ratio.
2. Find the first term \(a_1\) and common ratio \(r\).
3. If asked for a specific term, use \(a_n=a_1r^{n-1}\).
4. If asked for a finite sum, use \(S_n=\frac{a_1(1-r^n)}{1-r}\).
5. If asked for an infinite sum, first check \(|r|<1\), then use \(S_\infty=\frac{a_1}{1-r}\).

10. Summary

A geometric sequence is a sequence where each term is found by multiplying by the same constant ratio. Its recursive form is \(a_n=ra_{n-1}\), and its explicit form is \(a_n=a_1r^{n-1}\).

A geometric series is the sum of the terms of a geometric sequence. For the first \(n\) terms, use

$$S_n = \frac{a_1(1-r^n)}{1-r}$$

when \(r\ne 1\). For an infinite geometric series, a finite sum exists only when \(|r|<1\), and then

$$S_\infty = \frac{a_1}{1-r}$$

If you can identify the first term and common ratio, you can solve most problems involving geometric sequences and series.

Infinite Geometric Series and Convergence

When we add terms of a sequence, we get a series. Some series stop after a certain number of terms, but others continue forever. A series that has infinitely many terms is called an infinite series.

In this lesson, we will focus on a special kind of infinite series called an infinite geometric series. You will learn how to recognize one, decide whether it converges or diverges, and find its exact sum when it does converge.

1. What is a geometric series?

A geometric sequence is a sequence where each term is found by multiplying the previous term by the same number. That number is called the common ratio, written as \(r\).

For example, in the sequence

$$3,\; 1.5,\; 0.75,\; 0.375,\; \dots$$

each term is multiplied by \(0.5\), so the common ratio is \(r = 0.5\).

If we add the terms of a geometric sequence, we get a geometric series. An infinite geometric series has the form

$$a + ar + ar^2 + ar^3 + \dots$$

where:

• \(a\) is the first term
• \(r\) is the common ratio

2. What does convergence mean?

An infinite series converges if the sum of more and more terms gets closer and closer to one fixed number. That fixed number is called the limit or the sum to infinity.

An infinite series diverges if its sum does not approach a single finite number.

For example, consider

$$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$$

The partial sums are:

• First term: \(1\)
• First two terms: \(1 + \frac{1}{2} = 1.5\)
• First three terms: \(1.75\)
• First four terms: \(1.875\)

These sums get closer and closer to \(2\), so the series converges to \(2\).

3. When does an infinite geometric series converge?

The infinite geometric series

$$a + ar + ar^2 + ar^3 + \dots$$

converges only when

$$|r| < 1$$

This means the common ratio must be between \(-1\) and \(1\):

$$-1 < r < 1$$

If \(|r| < 1\), the terms become smaller and smaller, approaching \(0\). Then the sum can settle toward a finite number.

If \(|r| \geq 1\), the terms do not shrink to \(0\), so the series does not converge.

Important cases:

• If \(r = 1\), the terms stay the same size, so the sum grows forever.
• If \(r = -1\), the terms alternate without settling to one value.
• If \(|r| > 1\), the terms get larger in size, so the series diverges.

4. Formula for the sum to infinity

If \(|r| < 1\), then the sum of the infinite geometric series is

$$S_{\infty} = \frac{a}{1-r}$$

This formula only works when the series converges, so always check that \(|r| < 1\) first.

5. Why does this formula work?

For a finite geometric series, the sum of the first \(n\) terms is

$$S_n = \frac{a(1-r^n)}{1-r} \quad \text{for } r \ne 1$$

If \(|r| < 1\), then as \(n\) gets very large, \(r^n \to 0\). So:

$$S_n = \frac{a(1-r^n)}{1-r} \to \frac{a(1-0)}{1-r} = \frac{a}{1-r}$$

That is why the infinite sum formula works.

6. How to solve problems involving infinite geometric series

1. Identify the first term \(a\).
2. Find the common ratio \(r\).
3. Check whether \(|r| < 1\).
4. If yes, use $$S_{\infty} = \frac{a}{1-r}$$
5. If no, state that the series diverges.

Worked Example 1: Basic converging series

Find the sum of

$$6 + 3 + 1.5 + 0.75 + \dots$$

Step 1: Identify \(a\) and \(r\)

• First term: \(a = 6\)
• Common ratio: \(r = \frac{3}{6} = 0.5\)

Step 2: Check convergence

Since \(|0.5| < 1\), the series converges.

Step 3: Use the formula

$$S_{\infty} = \frac{a}{1-r} = \frac{6}{1-0.5} = \frac{6}{0.5} = 12$$

Answer: The series converges to \(12\).

Worked Example 2: Fraction ratio

Determine whether the series converges, and if it does, find its sum:

$$8 - 4 + 2 - 1 + \dots$$

Step 1: Identify \(a\) and \(r\)

• First term: \(a = 8\)
• Common ratio: \(r = \frac{-4}{8} = -\frac{1}{2}\)

Step 2: Check convergence

$$\left|-\frac{1}{2}\right| = \frac{1}{2} < 1$$

So the series converges.

Step 3: Find the sum

$$S_{\infty} = \frac{8}{1-\left(-\frac{1}{2}\right)} = \frac{8}{1+\frac{1}{2}} = \frac{8}{\frac{3}{2}} = \frac{16}{3}$$

Answer: The series converges to \(\frac{16}{3}\).

Notice that the terms alternate signs, but the series still converges because the size of the terms keeps shrinking and \(|r| < 1\).

Worked Example 3: A divergent series

Determine whether the series converges:

$$5 + 10 + 20 + 40 + \dots$$

Step 1: Identify \(a\) and \(r\)

• \(a = 5\)
• \(r = \frac{10}{5} = 2\)

Step 2: Check convergence

Since \(|2| = 2 \geq 1\), the series does not converge.

Answer: The series diverges.

There is no finite sum to infinity because the terms are getting larger, not smaller.

Worked Example 4: Finding a missing value

An infinite geometric series has first term \(a = 15\) and sum to infinity \(25\). Find the common ratio \(r\).

Step 1: Use the formula

$$S_{\infty} = \frac{a}{1-r}$$

Substitute the known values:

$$25 = \frac{15}{1-r}$$

Step 2: Solve for \(r\)

Multiply both sides by \(1-r\):

$$25(1-r) = 15$$

$$25 - 25r = 15$$

$$-25r = -10$$

$$r = \frac{10}{25} = \frac{2}{5}$$

Step 3: Check convergence

Since \(\left|\frac{2}{5}\right| < 1\), this value is valid.

Answer: The common ratio is \(\frac{2}{5}\).

7. Common mistakes to avoid

• Using the sum to infinity formula when \(|r| \geq 1\): The formula only works for converging geometric series.
• Mixing up the first term and the common ratio: Always identify \(a\) as the first term and find \(r\) by dividing a term by the one before it.
• Forgetting absolute value: A ratio like \(-0.8\) still gives convergence because \(|-0.8| = 0.8 < 1\).
• Thinking alternating signs always mean divergence: An alternating geometric series can converge if \(|r| < 1\).

8. Quick check questions

Try these on your own:

1. Does $$4 + 2 + 1 + 0.5 + \dots$$ converge? If so, find the sum.
2. Does $$7 - 7 + 7 - 7 + \dots$$ converge?
3. Find the sum to infinity of $$\frac{3}{2} + \frac{3}{4} + \frac{3}{8} + \frac{3}{16} + \dots$$

Answers:

1. Yes. \(a=4\), \(r=\frac{1}{2}\), so $$S_{\infty} = \frac{4}{1-\frac{1}{2}} = 8$$
2. No. Here \(r=-1\), and since \(|r|=1\), the series diverges.
3. \(a=\frac{3}{2}\), \(r=\frac{1}{2}\), so $$S_{\infty} = \frac{\frac{3}{2}}{1-\frac{1}{2}} = 3$$

9. Summary

An infinite geometric series has the form $$a + ar + ar^2 + ar^3 + \dots$$. It converges only when the common ratio satisfies $$|r| < 1$$.

When it converges, its sum to infinity is

$$S_{\infty} = \frac{a}{1-r}$$

Always check the value of \(r\) before using the formula. If \(|r| \geq 1\), the series diverges and does not have a finite sum.

Sigma Notation and Sum Properties

When a series has many terms, writing every term can be long and difficult to read. Sigma notation gives us a compact way to represent a sum using the Greek capital letter sigma, written as \(\Sigma\).

This notation is especially useful in sequences and series because it helps us describe patterns, calculate totals, and apply algebra rules more efficiently.

In this lesson, you will learn how to read sigma notation, write sums in sigma form, expand sigma expressions into full sums, and use important sum properties to simplify expressions.

1. What Sigma Notation Means

A sum written in sigma notation has the general form

Each part has a meaning:

• \(\Sigma\) means “add.”
• \(i\) is the index, or counting variable.
• \(m\) is the starting value of the index.
• \(n\) is the ending value of the index.
• \(a_i\) is the rule for each term.

For example,

means substitute \(i=1,2,3,4\) and add the results:

So sigma notation is just a shorter way to write a repeated addition.

2. How to Expand a Sigma Expression

To expand a sigma expression:

1. Start with the lower value of the index.
2. Substitute each integer value up to the upper value.
3. Add the resulting terms.

Example:

Substitute \(k=2,3,4,5\):

This shows that the index simply tells us which values to plug into the term formula.

3. How to Write a Sum in Sigma Notation

To write a series using sigma notation, identify:

• the pattern of the terms,
• the variable used to count the terms,
• where the counting starts and ends.

For example, consider

These terms follow the rule \(3n\), starting at \(n=1\) and ending at \(n=5\). So we can write

Another example is

This can be written as

4. Important Sum Properties

Sigma notation follows several algebra rules. These help us break apart sums and simplify expressions.

Property 1: Sum of a constant multiple

This means a constant factor can be pulled outside the summation.

Example:

Property 2: Sum of a sum

This means we can split a sum of two expressions into two separate sums.

Example:

Property 3: Sum of a difference

Property 4: Sum of a constant

If \(c\) is a constant, then adding it repeatedly from \(i=m\) to \(i=n\) gives

This works because the number of terms from \(m\) to \(n\), inclusive, is

For example,

5. Common Summation Formulas

Some sums appear often enough that they have standard formulas. These are useful in 12th Grade Maths.

First formula: Sum of the first \(n\) positive integers

Example:

Second formula: Sum of squares

Third formula: Sum of cubes

You do not need to memorize every possible formula, but these three are very common and important.

6. Worked Examples

Example 1: Expand and evaluate a basic sigma expression

Evaluate

Step 1: Expand the sum.

Step 2: Simplify each term.

Step 3: Add.

Answer: \(25\)

Example 2: Write a series in sigma notation

Write the series

in sigma notation.

Step 1: Find the pattern.

This is an arithmetic pattern with first term 4 and common difference 3.

Step 2: Write a formula for the terms.

If we start indexing at \(n=1\), the term formula is

Check:

• When \(n=1\), \(3(1)+1=4\)
• When \(n=2\), \(3(2)+1=7\)
• When \(n=5\), \(3(5)+1=16\)

Step 3: Write the sum.

Answer: $$\sum_{n=1}^{5} (3n+1)$$

Example 3: Use sum properties to simplify

Evaluate

Step 1: Split the sum using the sum properties.

Step 2: Evaluate each part.

First,

So,

Next,

There are 4 terms because \(k\) goes from 1 to 4.

Step 3: Subtract.

Answer: \(48\)

Example 4: Use a standard summation formula

Evaluate

Step 1: Split the sum.

Step 2: Use formulas.

Also,

Step 3: Combine the results.

Answer: \(710\)

7. Common Mistakes to Avoid

• Forgetting to include both endpoints. In \(\sum_{i=1}^{5}\), the values are \(1,2,3,4,5\), not just \(1\) through \(4\).
• Using the wrong term formula. Always test your rule with the first few terms.
• Confusing the index with the value of the term. The index is just the counting number used in the formula.
• Forgetting how many times a constant is added. For \(\sum_{i=m}^{n} c\), there are \(n-m+1\) terms.
• Not using brackets carefully. In expressions like \(\sum (2i+3)\), both parts are included in each term.

8. Why Sigma Notation Matters

Sigma notation is more than just shorthand. It helps connect sequences, series, algebra, and later topics in mathematics. It allows us to describe long patterns clearly, use formulas efficiently, and solve problems that would be messy if every term had to be written out.

It is also useful in combinatorics and discrete mathematics, where counting patterns often involve repeated addition.

Brief Summary

Sigma notation uses the symbol \(\Sigma\) to write sums in a compact form. You can expand sigma notation by substituting values of the index, and you can write a series in sigma form by identifying the term rule and the first and last index values.

The main sum properties let you pull out constants and split sums across addition or subtraction. Standard formulas such as

make many calculations much faster. With practice, sigma notation becomes a powerful tool for working with sequences and series.

Proof by Mathematical Induction is a method used to prove that a mathematical statement is true for all natural numbers, usually starting from \(n=1\) or another starting value like \(n=0\).

It is especially useful in topics such as sequences, series, formulas, divisibility, and combinatorics, where a pattern depends on a whole number \(n\).

The main idea is like a row of falling dominoes:

• First, you show the first domino falls. This is the base case.
• Then, you show that if one domino falls, the next one must also fall. This is the inductive step.
• So, all the dominoes fall. In maths, that means the statement is true for every natural number from the starting point onward.

This lesson will show you exactly how induction works, when to use it, and how to write a clear proof.

1. When do we use mathematical induction?

We use induction when we have a statement involving \(n\), and we want to prove it is true for every natural number.

Some common examples are:

• sum formulas, such as \(1+2+3+\cdots+n = \frac{n(n+1)}{2}\)
• formulas for sequences
• divisibility statements, such as showing an expression is always divisible by a certain number
• inequalities that hold for all \(n\ge 1\)

2. The structure of an induction proof

Suppose we want to prove a statement \(P(n)\) for all \(n \ge 1\).

A proof by induction has two essential parts.

1. Base Case: Show that \(P(1)\) is true.
2. Inductive Step:
   • Assume \(P(k)\) is true for some natural number \(k\). This is called the induction hypothesis.
   • Using that assumption, prove that \(P(k+1)\) is true.

If both parts are correct, then \(P(n)\) is true for all natural numbers \(n\ge 1\).

3. Important language in induction

• Statement \(P(n)\): the claim you want to prove.
• Base case: the first value where the statement begins.
• Induction hypothesis: the temporary assumption that \(P(k)\) is true.
• Inductive conclusion: proving \(P(k+1)\) using the hypothesis.

A very important point is this: in the inductive step, you do not assume the thing you are trying to prove for \(k+1\). You only assume it is true for \(k\), and then use that to build the next case.

4. General template for writing an induction proof

Here is a standard format you can follow.

To prove: \(P(n)\) is true for all \(n\ge 1\).

1. Base case: Verify \(P(1)\).
2. Inductive hypothesis: Assume \(P(k)\) is true for some \(k\ge 1\).
3. Inductive step: Show that this assumption implies \(P(k+1)\).
4. Conclusion: Therefore, by mathematical induction, \(P(n)\) is true for all \(n\ge 1\).

Now let us work through examples from easier to harder.

Worked Example 1: Sum of the first \(n\) natural numbers

Prove that

for all natural numbers \(n\ge 1\).

Step 1: Define the statement

Let

Step 2: Base case

Check \(n=1\):

Left side: \(1\)

Right side: \(\frac{1(1+1)}{2} = \frac{2}{2}=1\)

So \(P(1)\) is true.

Step 3: Induction hypothesis

Assume that \(P(k)\) is true for some \(k\ge 1\). That means:

Step 4: Prove \(P(k+1)\)

We must show:

Start with the left side of \(P(k+1)\):

Here we used the induction hypothesis.

Now simplify:

This is exactly the formula for \(P(k+1)\).

Conclusion: By mathematical induction,

for all \(n\ge 1\).

What to notice:

• We checked the first case carefully.
• We assumed the formula for \(k\), not for \(k+1\).
• We replaced the sum up to \(k\) using the induction hypothesis, then simplified.

Worked Example 2: Sum of the first \(n\) odd numbers

Prove that

for all \(n\ge 1\).

Step 1: Define the statement

Step 2: Base case

For \(n=1\):

Left side: \(1\)

Right side: \(1^2=1\)

So \(P(1)\) is true.

Step 3: Induction hypothesis

Assume \(P(k)\) is true for some \(k\ge 1\). Then:

Step 4: Prove \(P(k+1)\)

The next odd number after \(2k-1\) is

So the left side of \(P(k+1)\) is:

Using the induction hypothesis:

This is exactly the right side of \(P(k+1)\).

Conclusion: By mathematical induction,

for all \(n\ge 1\).

Why this works: each new case is formed by taking the previous sum and adding the next term. That makes induction a natural method.

Worked Example 3: A divisibility proof

Prove that \(7^n-1\) is divisible by \(6\) for all natural numbers \(n\ge 1\).

Step 1: Define the statement

Let \(P(n)\) be: \(7^n-1\) is divisible by \(6\).

This means we want to show that \(7^n-1 = 6m\) for some integer \(m\).

Step 2: Base case

For \(n=1\):

Since \(6\) is divisible by \(6\), the base case is true.

Step 3: Induction hypothesis

Assume that for some \(k\ge 1\), \(7^k-1\) is divisible by \(6\).

So there exists an integer \(m\) such that

Step 4: Prove \(P(k+1)\)

Consider:

Now rewrite this in a useful way:

Using the induction hypothesis \(7^k-1=6m\):

So \(7^{k+1}-1\) is divisible by \(6\).

Conclusion: By mathematical induction, \(7^n-1\) is divisible by \(6\) for all \(n\ge 1\).

What to learn from this example:

• In divisibility proofs, it often helps to rewrite the expression so the induction hypothesis appears inside it.
• Then factor out the required number.

Worked Example 4: A combinatorics-style formula

Prove that

for all natural numbers \(n\ge 1\).

This kind of statement appears often in discrete maths and combinatorics, where powers of \(2\) count choices or subsets.

Step 1: Define the statement

Step 2: Base case

For \(n=1\):

So \(P(1)\) is true.

Step 3: Induction hypothesis

Assume that for some \(k\ge 1\),

Step 4: Prove \(P(k+1)\)

We want to show:

Start with:

Since \(2^k > k\), multiplying both sides by \(2\) gives

Now for all \(k\ge 1\), we know

because \(2k-(k+1)=k-1\ge 0\).

So:

Therefore,

So \(P(k+1)\) is true.

Conclusion: By mathematical induction,

for all \(n\ge 1\).

5. How to think during the inductive step

Many students understand the base case but get stuck on the inductive step. A good strategy is to ask:

• What does the statement look like for \(k\)?
• What does it need to look like for \(k+1\)?
• How can I start from the \(k+1\) version and use the \(k\) version inside it?

For sums, this usually means separating off the last term.

For example:

can be seen as

Then the bracket can be replaced using the induction hypothesis.

For divisibility, try rewriting the new expression so that the old expression appears inside it.

For inequalities, use the induction hypothesis and then compare carefully to the new target.

6. Common mistakes to avoid

• Skipping the base case. Without the first true case, the proof is incomplete.
• Using the wrong assumption. You may assume \(P(k)\), not \(P(k+1)\).
• Not stating the induction hypothesis clearly. Write exactly what you are assuming.
• Doing algebra too quickly. Show enough steps so the logic is clear.
• Forgetting the conclusion. End by saying the statement is true for all required natural numbers by induction.

7. A checklist for every induction proof

• Have I clearly written the statement \(P(n)\)?
• Did I verify the base case correctly?
• Did I state the induction hypothesis for some \(k\)?
• Did I use the hypothesis to prove the statement for \(k+1\)?
• Did I write a final conclusion?

8. Why induction is important in sequences, series, and combinatorics

In 12th Grade Maths, many formulas depend on counting patterns that continue step by step.

• In sequences, induction can prove formulas for the \(n\)th term.
• In series, induction can prove formulas for sums of the first \(n\) terms.
• In combinatorics, induction can prove counting results that grow from one case to the next.

So induction is not just a proof technique. It is a way of showing that a pattern truly continues forever, not just for the first few values.

Brief Summary

Proof by mathematical induction is used to show that a statement is true for all natural numbers.

You first prove the base case, then assume the statement is true for \(k\), and use that assumption to prove it is true for \(k+1\).

If both parts work, then the statement is true for every natural number from the starting value onward.

When writing induction proofs, be careful to state your assumption clearly, use it correctly, and finish with a conclusion.

The Fundamental Counting Principle is one of the most important ideas in combinatorics. It helps us count the total number of possible outcomes in a process that happens in stages.

If one stage can happen in several ways, and the next stage can also happen in several ways, then the total number of outcomes is found by multiplying the number of choices at each stage.

This principle is especially useful when listing every possibility would take too long. Instead of writing all outcomes one by one, we can count them efficiently.

Main Idea

If an event has:

• Stage 1 with \(a\) choices,
• Stage 2 with \(b\) choices,
• Stage 3 with \(c\) choices,

then the total number of possible outcomes is

$$a \times b \times c$$

More generally, if there are several independent stages, multiply the number of choices in each stage.

Why Multiplication Works

Suppose you are choosing an outfit. If you have 3 shirts and 2 pairs of pants, then for each shirt, there are 2 pants you could wear.

So the total number of outfits is

$$3 \times 2 = 6$$

This works because each choice in the first stage can be paired with every choice in the second stage.

When to Use the Fundamental Counting Principle

Use this principle when:

• a task happens in separate steps or stages,
• you know how many choices are available at each stage,
• you want the total number of complete outcomes.

It is often used in problems involving:

• forming passwords or codes,
• choosing outfits, meals, or schedules,
• arranging selections step by step,
• counting paths or combinations of decisions.

Important Idea: Independent Stages

The simplest use of the Fundamental Counting Principle is when the stages are independent. This means the number of choices at one stage does not change because of earlier choices.

For example, if a restaurant offers 4 drinks and 5 sandwiches, then choosing a drink does not change the number of sandwich options. The total number of meals is

$$4 \times 5 = 20$$

What if the Number of Choices Changes?

Sometimes the stages are not fully independent. For example, if you are choosing 3 different students from a group, then after choosing the first student, there is one fewer choice left for the second stage.

You can still use the Fundamental Counting Principle. You just multiply the number of choices available at each stage as it happens.

For example, choosing 3 different students from 10 students in order gives:

$$10 \times 9 \times 8 = 720$$

So the principle still works even when the number of choices changes from stage to stage.

Worked Example 1: Simple Two-Stage Event

A cafeteria offers 3 types of sandwiches and 4 types of drinks. How many different lunch combinations can a student choose if they pick one sandwich and one drink?

Step 1: Count the choices for each stage.

• Sandwiches: \(3\)
• Drinks: \(4\)

Step 2: Multiply.

$$3 \times 4 = 12$$

Answer: There are 12 possible lunch combinations.

Worked Example 2: Three-Stage Event

A car can be ordered in 5 colors, with 2 engine types, and 3 interior styles. How many different cars can be made?

Step 1: Identify the stages.

• Color: \(5\) choices
• Engine: \(2\) choices
• Interior: \(3\) choices

Step 2: Multiply all the choices.

$$5 \times 2 \times 3 = 30$$

Answer: There are 30 different cars possible.

Worked Example 3: Passwords

A password consists of 2 letters followed by 3 digits. Assume letters can be any of the 26 letters of the alphabet and digits can be any of the 10 digits from 0 to 9. How many passwords are possible if repetition is allowed?

Step 1: Break the password into stages.

• First letter: \(26\) choices
• Second letter: \(26\) choices
• First digit: \(10\) choices
• Second digit: \(10\) choices
• Third digit: \(10\) choices

Step 2: Multiply.

$$26 \times 26 \times 10 \times 10 \times 10$$

$$= 26^2 \times 10^3$$

$$= 676000$$

Answer: There are 676,000 possible passwords.

Worked Example 4: No Repetition Allowed

How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 if no digit may be repeated?

Step 1: Count choices for each digit position.

• First digit: 5 choices
• Second digit: 4 choices
• Third digit: 3 choices

The number of choices decreases because digits cannot repeat.

Step 2: Multiply.

$$5 \times 4 \times 3 = 60$$

Answer: There are 60 different 3-digit numbers.

Common Mistakes to Avoid

• Adding instead of multiplying: If the problem has stages that happen together to make one outcome, multiply. Adding is used when choosing between separate alternatives, not combining stages.
• Forgetting restrictions: If repetition is not allowed, the number of choices may decrease at each stage.
• Missing a stage: Be sure to include every step in the process.
• Using the same number of choices each time when you should not: Check whether earlier choices affect later ones.

How to Recognize These Problems

Ask yourself these questions:

1. Is the task completed in more than one step?
2. Do I know the number of choices in each step?
3. Am I creating one complete outcome by making all of these choices?

If the answer is yes, then the Fundamental Counting Principle is probably the right method.

Quick Comparison: Multiply or Add?

It is helpful to know when to multiply and when to add.

• Multiply when one outcome is made by completing several stages.
• Add when you are choosing from separate options that do not happen together.

For example:

• Choose 1 shirt and 1 pair of pants: multiply.
• Choose either a sandwich or a salad: add.

Practice Thinking

Suppose a school ID code uses 1 letter followed by 4 digits, and repetition is allowed. The number of possible codes is:

$$26 \times 10 \times 10 \times 10 \times 10 = 26 \times 10^4 = 260000$$

This shows how quickly the number of outcomes grows when there are several stages.

Summary

The Fundamental Counting Principle says that if a process has several stages, the total number of possible outcomes is found by multiplying the number of choices at each stage.

It works for independent stages and also for situations where the number of choices changes, as long as you count the available choices step by step.

This principle is a foundation for many topics in combinatorics, including arrangements, codes, selections, and more advanced counting methods.

Permutations and Combinations are counting methods used in combinatorics. They help us answer questions like: “How many different ways can items be arranged?” or “How many different groups can be chosen?”

The most important idea is this: if order matters, use permutations. If order does not matter, use combinations.

For example, choosing a class president, vice-president, and secretary is different from simply choosing 3 students for a committee. In the first case, the jobs are different, so order matters. In the second case, a group of 3 students is just a group, so order does not matter.

To understand these ideas well, we first need to review factorials.

A factorial is written with an exclamation mark. For a positive integer \(n\),

Some examples are:

• \(1! = 1\)
• \(3! = 3\cdot 2\cdot 1 = 6\)
• \(5! = 5\cdot 4\cdot 3\cdot 2\cdot 1 = 120\)

Also, by definition,

This is important in many counting formulas.

Why factorials appear in counting: If 5 different books are arranged on a shelf, there are 5 choices for the first spot, 4 for the second, 3 for the third, 2 for the fourth, and 1 for the last. By the multiplication principle, the total number of arrangements is

This leads directly to permutations.

Permutations count ordered arrangements. If we want to arrange \(r\) objects chosen from \(n\) different objects, the formula is

Another common notation is \(P(n,r)\). Both mean the same thing.

Why does this formula work? Suppose we want to fill \(r\) positions from \(n\) objects.

• There are \(n\) choices for the first position.
• Then \(n-1\) choices for the second.
• Then \(n-2\) choices for the third.
• This continues until \(r\) positions are filled.

So the total is

Combinations count unordered selections. If we want to choose \(r\) objects from \(n\) different objects, the formula is

Another common notation is \(C(n,r)\) or \(\binom{n}{r}\).

Why do we divide by \(r!\)? When we use permutations, we count every possible order of the same group. But in a combination, those different orders should count as only one selection.

For example, if we choose 3 letters A, B, and C, the permutations ABC, ACB, BAC, BCA, CAB, and CBA are 6 different orders. But as a combination, they are all the same group: {A, B, C}. Since \(3! = 6\), we divide by \(r!\) to remove the extra counting.

This gives the relationship:

Now let us look at how to decide whether a problem is a permutation or a combination.

• Use a permutation when positions, rankings, or roles matter.
• Use a combination when you are just selecting a group and the order does not matter.

Here are some quick examples:

• Selecting 3 winners for gold, silver, and bronze: permutation
• Choosing 3 students for a team: combination
• Making a 4-digit code from different digits: permutation
• Choosing 4 books to take on a trip: combination

Worked Example 1: Simple permutation

How many ways can 3 students be arranged in a line from a group of 7 students?

Because the students are being arranged in a line, order matters. So we use permutations:

Now simplify:

Answer: There are \(210\) ways.

Worked Example 2: Simple combination

How many ways can 4 players be chosen from 10 players for a basketball practice squad?

We are choosing 4 players, and the order does not matter. So we use combinations:

Simplify:

Answer: There are \(210\) possible squads.

Notice something interesting: in these two examples, the answers are the same, but the situations are different. This does not always happen. The key is to choose the correct method based on whether order matters.

Worked Example 3: Distinguishing order

A club has 8 members. In how many ways can it choose a president, vice-president, and treasurer?

These are different positions, so order matters. The person chosen as president is different from the person chosen as treasurer. Therefore, use permutations:

Answer: There are \(336\) ways.

Now compare that with this question: In how many ways can the same club choose a 3-person committee?

This time, all 3 members have the same role. Order does not matter, so use combinations:

Answer: There are \(56\) ways.

This example clearly shows the difference between permutations and combinations. The same numbers \(n=8\) and \(r=3\) give very different answers depending on whether order matters.

Worked Example 4: Using combinations in a word problem

From 12 books, a student wants to choose 5 books to take on vacation. How many different selections are possible?

The student is just selecting 5 books. The order of the books does not matter. So we use combinations:

Simplify by canceling:

Answer: There are \(792\) different selections.

Important properties of combinations

Combinations have a useful symmetry property:

This means choosing \(r\) objects from \(n\) is the same as leaving out \(n-r\) objects.

For example, choosing 2 students from 7 is the same as choosing which 5 students are not selected:

This property can make calculations easier.

Common mistakes to avoid

• Using permutations when order does not matter. If you are choosing a group, do not count different orders of the same group.
• Using combinations when order does matter. If positions, rankings, or arrangements are involved, use permutations.
• Not simplifying factorials correctly. Write out only as much of the factorial as needed to cancel.
• Forgetting that \(0! = 1\). This matters in some formulas.

Strategy for solving counting problems

1. Read the question carefully.
2. Ask: Does order matter?
3. If yes, use \(^nP_r\).
4. If no, use \(^nC_r\).
5. Substitute values carefully.
6. Simplify step by step.

Quick comparison

• Permutation: ordered arrangement
• Combination: unordered selection
• Permutation formula: \(^nP_r = \frac{n!}{(n-r)!}\)
• Combination formula: \(^nC_r = \frac{n!}{r!(n-r)!}\)

Summary

Permutations and combinations are both methods for counting. The main difference is whether order matters. If order matters, use permutations; if order does not matter, use combinations.

Factorials are the foundation of both formulas. Learning to identify the type of problem and apply the correct formula is the key skill. With practice, you will quickly recognize whether a situation involves arranging or simply choosing.

The Binomial Theorem and Pascal’s Triangle

When we expand expressions like \((x+2)^2\) or \((a+b)^3\), we multiply the same binomial by itself several times. For small powers, this is manageable by direct multiplication. But for larger powers such as \((x+3)^8\), multiplying again and again becomes slow and easy to get wrong.

The Binomial Theorem gives a fast, organized way to expand any expression of the form \((a+b)^n\). Pascal’s Triangle helps us find the coefficients in that expansion. Together, they are powerful tools in algebra and combinatorics.

This lesson will show you how Pascal’s Triangle is built, how it connects to binomial expansions, and how to use the Binomial Theorem to write terms in an expansion quickly and accurately.

1. What is a binomial?

A binomial is an expression with two terms, such as:

• \(x+1\)
• \(2a-3\)
• \(m+n\)

When a binomial is raised to a power, we get an expansion:

$$ (a+b)^n $$

For example:

• \((a+b)^1=a+b\)
• \((a+b)^2=a^2+2ab+b^2\)
• \((a+b)^3=a^3+3a^2b+3ab^2+b^3\)

Notice that the coefficients \(1,2,1\) and \(1,3,3,1\) follow a pattern. That pattern comes from Pascal’s Triangle.

2. Pascal’s Triangle

Pascal’s Triangle is an arrangement of numbers where:

• Each row starts and ends with \(1\).
• Every inside number is found by adding the two numbers directly above it.

The first few rows are:

$$ \begin{array}{ccccccccc} &&&&1&&&&\\ &&&1&&1&&&\\ &&1&&2&&1&&\\ &1&&3&&3&&1&\\ 1&&4&&6&&4&&1 \end{array} $$

If we continue, the next rows are:

$$ 1\quad 5\quad 10\quad 10\quad 5\quad 1 $$

and

$$ 1\quad 6\quad 15\quad 20\quad 15\quad 6\quad 1 $$

These rows match the coefficients of the expansions of \((a+b)^0, (a+b)^1, (a+b)^2, (a+b)^3\), and so on.

Connection between rows and powers

• Row \(0\): \(1\) gives \((a+b)^0\)
• Row \(1\): \(1,1\) gives \((a+b)^1\)
• Row \(2\): \(1,2,1\) gives \((a+b)^2\)
• Row \(3\): \(1,3,3,1\) gives \((a+b)^3\)
• Row \(4\): \(1,4,6,4,1\) gives \((a+b)^4\)

3. Patterns in binomial expansions

Consider these expansions:

$$ (a+b)^2=a^2+2ab+b^2 $$

$$ (a+b)^3=a^3+3a^2b+3ab^2+b^3 $$

$$ (a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4 $$

There are important patterns:

• The coefficients come from Pascal’s Triangle.
• The power of \(a\) starts at \(n\) and decreases by \(1\) each term.
• The power of \(b\) starts at \(0\) and increases by \(1\) each term.
• The sum of the exponents in each term is always \(n\).

For \((a+b)^4\), the terms are built like this:

• \(a^4b^0\)
• \(a^3b^1\)
• \(a^2b^2\)
• \(a^1b^3\)
• \(a^0b^4\)

Then we attach the coefficients \(1,4,6,4,1\).

4. The Binomial Theorem

The Binomial Theorem gives a general formula for expanding \((a+b)^n\):

$$ (a+b)^n=\sum_{r=0}^{n} {n \choose r} a^{n-r}b^r $$

Here, \({n \choose r}\) is called a binomial coefficient. It tells us the coefficient of the term containing \(a^{n-r}b^r\).

The binomial coefficient is calculated by:

$$ {n \choose r}=\frac{n!}{r!(n-r)!} $$

The symbol \(!\) means factorial. For example:

• \(4!=4\cdot3\cdot2\cdot1=24\)
• \(0!=1\)

So if you want the coefficient of a term, you can get it either:

• from Pascal’s Triangle, or
• by using \({n \choose r}\).

5. Reading the general term

The general term in the expansion of \((a+b)^n\) is:

$$ T_{r+1}={n \choose r}a^{n-r}b^r $$

This means:

• \(r\) starts at \(0\)
• the first term is \(T_1\)
• the second term is found using \(r=1\)
• the third term is found using \(r=2\), and so on

This is especially useful when you only need one particular term, not the whole expansion.

Worked Example 1: Expand \((x+2)^4\)

Step 1: Find the coefficients from Pascal’s Triangle for power \(4\):

$$ 1,4,6,4,1 $$

Step 2: Write the powers of \(x\) decreasing from \(4\) to \(0\):

$$ x^4, x^3, x^2, x, 1 $$

Step 3: Write the powers of \(2\) increasing from \(0\) to \(4\):

$$ 2^0,2^1,2^2,2^3,2^4 $$

Step 4: Combine everything:

$$ (x+2)^4=1(x^4)(2^0)+4(x^3)(2^1)+6(x^2)(2^2)+4(x)(2^3)+1(2^4) $$

Simplify each term:

$$ (x+2)^4=x^4+8x^3+24x^2+32x+16 $$

Worked Example 2: Expand \((2a-b)^3\)

This time the second term is negative, so we must be careful with signs.

For power \(3\), the Pascal coefficients are:

$$ 1,3,3,1 $$

Write the expansion pattern:

$$ (2a-b)^3=1(2a)^3+3(2a)^2(-b)+3(2a)(-b)^2+1(-b)^3 $$

Now simplify:

$$ (2a)^3=8a^3 $$

$$ 3(2a)^2(-b)=3(4a^2)(-b)=-12a^2b $$

$$ 3(2a)(-b)^2=3(2a)(b^2)=6ab^2 $$

$$ (-b)^3=-b^3 $$

So the expansion is:

$$ (2a-b)^3=8a^3-12a^2b+6ab^2-b^3 $$

Important note: odd powers of a negative stay negative, but even powers become positive.

Worked Example 3: Expand \((x+3)^5\)

For power \(5\), the Pascal coefficients are:

$$ 1,5,10,10,5,1 $$

Use the pattern:

$$ (x+3)^5=1(x^5)(3^0)+5(x^4)(3^1)+10(x^3)(3^2)+10(x^2)(3^3)+5(x)(3^4)+1(3^5) $$

Now simplify:

$$ (x+3)^5=x^5+15x^4+90x^3+270x^2+405x+243 $$

This example shows how much faster Pascal’s Triangle is than repeated multiplication.

Worked Example 4: Find the fourth term in \((2x-5)^6\)

We do not need the whole expansion. We use the general term:

$$ T_{r+1}={6 \choose r}(2x)^{6-r}(-5)^r $$

The fourth term is \(T_4\), so:

$$ r+1=4 \Rightarrow r=3 $$

Substitute \(r=3\):

$$ T_4={6 \choose 3}(2x)^{3}(-5)^3 $$

Now calculate each part:

$$ {6 \choose 3}=\frac{6!}{3!3!}=20 $$

$$ (2x)^3=8x^3 $$

$$ (-5)^3=-125 $$

Multiply:

$$ T_4=20\cdot 8x^3\cdot(-125) $$

$$ T_4=-20000x^3 $$

So the fourth term is:

$$ -20000x^3 $$

6. Using combinations to understand the coefficients

The coefficient \({n \choose r}\) counts how many ways to choose \(r\) factors of \(b\) from the \(n\) brackets in \((a+b)^n\).

For example, in:

$$ (a+b)^4=(a+b)(a+b)(a+b)(a+b) $$

To get a term with \(a^2b^2\), we must choose \(b\) from exactly 2 of the 4 brackets, and choose \(a\) from the other 2. The number of ways to do that is:

$$ {4 \choose 2}=6 $$

That is why the coefficient of \(a^2b^2\) in \((a+b)^4\) is \(6\).

This is the combinatorics connection in the Binomial Theorem: expansion coefficients count arrangements of choices.

7. Helpful steps for expanding any binomial

1. Identify the power \(n\).
2. Get the coefficients from Pascal’s Triangle or \({n \choose r}\).
3. Write the first term raised to descending powers.
4. Write the second term raised to ascending powers.
5. Multiply carefully, including signs and numerical powers.

8. Common mistakes to avoid

• Using the wrong row of Pascal’s Triangle: the power \(n\) matches row \(n\).
• Forgetting powers on numbers: in \((x+3)^5\), the 3 must also be raised to powers.
• Sign errors: in expressions like \((a-b)^n\), negative signs matter.
• Skipping terms: there should be \(n+1\) terms in \((a+b)^n\).
• Wrong exponents: one variable’s power decreases while the other increases.

9. Why this topic matters

The Binomial Theorem is important because it connects algebra and combinatorics. It helps in:

• expanding expressions quickly,
• finding specific terms without full expansion,
• understanding combinations,
• working with probability and counting problems.

Pascal’s Triangle is also useful beyond algebra. Its patterns appear in probability, number theory, and recursive structures.

Brief Summary

Pascal’s Triangle provides the coefficients for expansions of \((a+b)^n\). The Binomial Theorem gives the general formula:

$$ (a+b)^n=\sum_{r=0}^{n} {n \choose r}a^{n-r}b^r $$

In every expansion, the first term’s exponent decreases, the second term’s exponent increases, and the coefficients come from combinations. Once you recognize these patterns, you can expand high-power binomials and find specific terms much more efficiently.