# Binomial Theorem

How to solve the binomial theorem problems: formula, examples, and their solutions.

## Formula

To write (*x* + *y*)^{1},

you should choose either *x* or *y* 1 time.

So you can write (*x* + *y*) like this:

Ways of choosing *y* 0 times (= *x* 1 time):_{1}C_{0}⋅*x*^{1}⋅*y*^{1}

Ways of choosing *y* 1 time (= *x* 0 times):_{1}C_{1}⋅*x*^{0}⋅*y*^{1}

Combinations (_{n}C_{r})

To expand (*x* + *y*)^{2},

you should choose either *x* or *y* 2 times

from each (*x* + *y*),

and multiply the choosed factors.

Ways of choosing *y* 0 times (= *x* 2 times):_{2}C_{0}⋅*x*^{2}⋅*y*^{0}

Ways of choosing *y* 1 time (= *x* 1 time):_{2}C_{1}⋅*x*^{1}⋅*y*^{1}

Ways of choosing *y* 2 times (= *x* 0 times):_{2}C_{2}⋅*x*^{0}⋅*y*^{2}

The result is the same as using the FOIL method:*x*^{2} + 2*xy* + *y*^{2}

To expand (*x* + *y*)^{3},

you should choose either *x* or *y* 3 times

from each (*x* + *y*),

and multiply the choosed factors.

Ways of choosing *y* 0 times (= *x* 3 times):_{3}C_{0}⋅*x*^{3}⋅*y*^{0}

Ways of choosing *y* 1 time (= *x* 2 times):_{3}C_{1}⋅*x*^{2}⋅*y*^{1}

Ways of choosing *y* 2 times (= *x* 1 time):_{3}C_{2}⋅*x*^{1}⋅*y*^{2}

Ways of choosing *y* 3 times (= *x* 0 times):_{3}C_{3}⋅*x*^{0}⋅*y*^{3}

To expand (*x* + *y*)^{n},

you should choose either *x* or *y* *n* times

from each (*x* + *y*),

and multiply the choosed factors.

Ways of choosing *y* 0 times (= *x* *n* times):_{n}C_{0}⋅*x*^{n}⋅*y*^{0}

Ways of choosing *y* 1 time (= *x* '*n* - 1' times):_{n}C_{1}⋅*x*^{n - 1}⋅*y*^{1}

Ways of choosing *y* 2 times (= *x* '*n* - 2' times):_{n}C_{2}⋅*x*^{n - 2}⋅*y*^{2}

...

So (*x* + *y*)^{n} is the sum of _{n}C_{k}⋅*x*^{n - k}⋅*y*^{k}

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

Sigma Notation

This is the formula of the binomial theorem.

## Example 1

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

and multiply the choosed factors.

Ways of choosing 4*y* 0 times (= *x* 3 times):_{3}C_{0}⋅*x*^{3}⋅(4*y*)^{0}

Ways of choosing 4*y* 1 time (= *x* 2 times):_{3}C_{1}⋅*x*^{2}⋅(4*y*)^{1}

Ways of choosing 4*y* 2 times (= *x* 1 time):_{3}C_{2}⋅*x*^{1}⋅(4*y*)^{2}

Ways of choosing 4*y* 3 times (= *x* 0 times):_{3}C_{3}⋅*x*^{0}⋅(4*y*)^{3}

Combinations (_{n}C_{r})

## Example 2

Recall the sigma formula of (*a* - 2*b*)^{8}.*k* goes from 0 to 8:

0, 1, 2, 3, ... .

So the fourth term is when *k* = 3.

So write the ways of

choosing -2*b* 3 times (= *a* '8 - 3' times):_{8}C_{3}⋅*a*^{8 - 3}⋅(-2*b*)^{3}.