# Binomial Theorem

How to use the binomial theorem to expand (a + b)^{n}: formula, 2 examples, and their solutions.

## Formula

### Formula

To expand the power of a binomial (x + y)^{n}:

Write _{n}C_{0}.

Write x^{n - 0} = x^{n}.

And write y^{0}.

Write +_{n}C_{1}.

Write x^{n - 1}.

And write y^{1}.

Write +_{n}C_{2}.

Write x^{n - 2}.

And write y^{2}.

Repeat this from k = 0 to k = n.

This is the binomial theorem.

Sigma Notation

Meaning of _{n}C_{k}x^{n - k}y^{k}_{n}C_{k}:

The number of ways

to pick k of y from n trials

Combination

y^{k}: Multiply the picked y factors.

(k of them)

x^{n - k}: Multiply the picked x factors.

(n - k of them)

## Example 1

### Example

### Solution

(x + 4y)^{3} is the power of a binomial.

So use the binomial theorem

to expand this expression.

Case 1: Pick 0 of +4y.

Write,

the number of ways

to pick 0 of +4y from 3 trials,_{3}C_{0}

times x^{3 - 0} = x^{3}

times (4y)^{0}.

Plus...

Case 2: Pick 1 of +4y.

Write,

the number of ways

to pick 1 of +4y from 3 trials,_{3}C_{1}

times x^{3 - 1} = x^{2}

times (4y)^{1}.

Plus...

Case 3: Pick 2 of +4y.

Write,

the number of ways

to pick 2 of +4y from 3 trials,_{3}C_{2}

times x^{3 - 2} = x^{1}

times (4y)^{2}.

Plus...

Case 4: Pick 3 of +4y.

Write,

the number of ways

to pick 3 of +4y from 3 trials,_{3}C_{3}

times x^{3 - 3} = x^{0}

times (4y)^{3}.

So (x + 4y)^{3}

= _{3}C_{0}x^{3}(4y)^{0} + _{3}C_{1}x^{2}(4y)^{1} + _{3}C_{2}x^{1}(4y)^{2} + _{3}C_{3}x^{0}(4y)^{3}.

_{3}C_{0} = 1

(4y)^{0} = 1

Zero Exponent_{3}C_{1} = 3/1 = 3

(4y)^{1} = 4y_{3}C_{2} = _{3}C_{1}

x^{1} = x

(4y)^{2} = 4^{2}y^{2} = 16y^{2}_{3}C_{3} = 1

x^{0} = 1

(4y)^{3} = 4^{3}y^{3} = 64y^{3}

Power of a Product

1⋅x^{3}⋅1 = x^{3}

+3⋅x^{2}⋅4y = +12x^{2}y

+_{3}C_{1} = +3

x⋅16y^{2} = 16xy^{2}

+1⋅1⋅64y^{3} = +64y^{3}

+3⋅16xy^{2} = +48xy^{2}

So

x^{3} + 12x^{2}y + 48xy^{2} + 64y^{3}

is the answer.

## Example 2

### Example

### Solution

The exponent of (a - 2b)^{8} is 8.

So k goes from 0 to 8.

It says

find the fourth term.

And it says

the first term is a^{8}.

So k is,

start from 0 and count 4:

0, 1, 2, and 3,

3.

So the fourth term is

when k = 3.

So write,

the number of ways

to pick 3 of -2b from 8 trials,_{8}C_{3}

times a^{8 - 3} = a^{5}

times (-2b)^{3}.

_{8}C_{3} = [8⋅7⋅6]/[3⋅2⋅1]

Combination

(-2b)^{3}

= (-2)^{3}⋅b^{3}

= (-8)⋅b^{3}

Cancel the denominator 3⋅2⋅1

and cancel the 6 in the numerator.

Then [8⋅7⋅6]/[3⋅2⋅1] = 8⋅7.

8⋅7⋅(-8)

= 56⋅(-8)

= -448

So -448a^{5}b^{3} is the answer.