# Binomial Theorem

How to solve the power of a sum, (*x* + *y*)^{n}, by using the binomial theorem: formula, examples, and their solutions.

## Formula

By using the binomial theorem,

you can expand (*x* + *y*)^{n}.

(*x* + *y*)^{n}

= _{n}C_{0}⋅*x*^{n}⋅*y*^{0} + _{n}C_{1}⋅*x*^{n - 1}⋅*y*^{1} + _{n}C_{2}⋅*x*^{n - 2}⋅*y*^{2} + ... + _{n}C_{n}⋅*x*^{0}⋅*y*^{n}.

Combinations (_{n}C_{r})

There's a pattern between the terms.

So you can write this in sigma notation:

[the sum of _{n}C_{k}⋅*x*^{n - k}⋅*y*^{k}

as *k* goes from 0 to *n*].

Sigma Notation

## Example 1: Expand (*x* + 4*y*)^{3}

Choose either *x* or 4*y* 3 times

and multiply the chosen factors.

The *x*^{3} term is,

choose *x* 3 times and (+4*y*) 0 times,_{3}C_{0}⋅*x*^{3}⋅(4*y*)^{0}.

The *x*^{2} term is,

choose *x* 2 times and (+4*y*) 1 time,_{3}C_{1}⋅*x*^{2}⋅(4*y*)^{1}.

The *x*^{1} term is,

choose *x* 1 time and (+4*y*) 2 times,_{3}C_{2}⋅*x*^{1}⋅(4*y*)^{2}.

The constant terms, *x*^{0} term, is,

choose *x* 0 times and (+4*y*) 3 times,_{3}C_{3}⋅*x*^{0}⋅(4*y*)^{3}.

_{3}C_{0} = 1

(4*y*)^{0} = 1_{3}C_{1} = 3_{3}C_{2} = _{3}C_{1}

= 3

(4*y*)^{2} = 16*y*^{2}_{3}C_{3} = _{3}C_{0}

= 1*x*^{0} = 1

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

+3⋅*x*^{2}⋅4*y* = +12*x*^{2}*y*

+3⋅*x*⋅16*y*^{2} = +48*x**y*^{2}

So (given) = *x*^{3} + 12*x*^{2}*y* + 48*x**y*^{2} + 64*y*^{3}.

## Example 2: Fourth Term of (*a* - 2*b*)^{8}

The exponent *n* is 8.

So *k* goes from 0 to 8.

So the fourth term is

when *k* is, (0, 1, 2, 3), [3].

So choose the later term, -2*b*,

[3] times.

And choose the former term, *a*,

8 - 3 = [5] times.

So the fourth term is_{8}C_{3}⋅*a*^{5}⋅(-2*b*)^{3}.

_{8}C_{3} = (8⋅7⋅6)/(3⋅2⋅1)

Combinations (_{n}C_{r})

(-2*b*)^{3} = -8*b*^{3}

Cancel the numerator 6

and cancel 3⋅2.

Then (8⋅7⋅6)/(3⋅2⋅1) = 56.

56⋅8 = 448

So the fourth term is -448*a*^{5}*b*^{3}.