Binomial Theorem

Binomial Theorem

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

Formula

(x + y)^n = [the sum of 'n_C_k * x^(n - k) * y^k' as k goes from 0 to n]

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):
1C0x1y1

Ways of choosing y 1 time (= x 0 times):
1C1x0y1

Combinations (nCr)

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):
2C0x2y0

Ways of choosing y 1 time (= x 1 time):
2C1x1y1

Ways of choosing y 2 times (= x 0 times):
2C2x0y2

The result is the same as using the FOIL method:
x2 + 2xy + y2

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):
3C0x3y0

Ways of choosing y 1 time (= x 2 times):
3C1x2y1

Ways of choosing y 2 times (= x 1 time):
3C2x1y2

Ways of choosing y 3 times (= x 0 times):
3C3x0y3

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):
nC0xny0

Ways of choosing y 1 time (= x 'n - 1' times):
nC1xn - 1y1

Ways of choosing y 2 times (= x 'n - 2' times):
nC2xn - 2y2
...

So (x + y)n is the sum of nCkxn - kyk
as k goes from 0 to n.

Sigma Notation

This is the formula of the binomial theorem.

Example 1

Expand the given expression. (x + 4y)^3

Choose either x or 4y 3 times
and multiply the choosed factors.

Ways of choosing 4y 0 times (= x 3 times):
3C0x3⋅(4y)0

Ways of choosing 4y 1 time (= x 2 times):
3C1x2⋅(4y)1

Ways of choosing 4y 2 times (= x 1 time):
3C2x1⋅(4y)2

Ways of choosing 4y 3 times (= x 0 times):
3C3x0⋅(4y)3

Combinations (nCr)

Example 2

Find the fourth term of the given expression. (a - 2b)^8

Recall the sigma formula of (a - 2b)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 -2b 3 times (= a '8 - 3' times):
8C3a8 - 3⋅(-2b)3.