Binomial Theorem

Binomial Theorem

How to solve the power of a sum, (x + y)n, by using the binomial theorem: 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]

By using the binomial theorem,
you can expand (x + y)n.

(x + y)n
= nC0xny0 + nC1xn - 1y1 + nC2xn - 2y2 + ... + nCnx0yn.

Combinations (nCr)

There's a pattern between the terms.

So you can write this in sigma notation:
[the sum of nCkxn - kyk
as k goes from 0 to n].

Sigma Notation





Example 1: Expand (x + 4y)3

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

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

The x3 term is,
choose x 3 times and (+4y) 0 times,
3C0x3⋅(4y)0.

The x2 term is,
choose x 2 times and (+4y) 1 time,
3C1x2⋅(4y)1.

The x1 term is,
choose x 1 time and (+4y) 2 times,
3C2x1⋅(4y)2.

The constant terms, x0 term, is,
choose x 0 times and (+4y) 3 times,
3C3x0⋅(4y)3.

3C0 = 1
(4y)0 = 1

3C1 = 3

3C2 = 3C1
= 3
(4y)2 = 16y2

3C3 = 3C0
= 1
x0 = 1
(4y)3 = 64y3

+3⋅x2⋅4y = +12x2y

+3⋅x⋅16y2 = +48xy2

So (given) = x3 + 12x2y + 48xy2 + 64y3.

Example 2: Fourth Term of (a - 2b)8

Find the fourth term of the given expression. (a - 2b)^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, -2b,
[3] times.

And choose the former term, a,
8 - 3 = [5] times.

So the fourth term is
8C3a5⋅(-2b)3.

8C3 = (8⋅7⋅6)/(3⋅2⋅1)

Combinations (nCr)

(-2b)3 = -8b3

Cancel the numerator 6
and cancel 3⋅2.

Then (8⋅7⋅6)/(3⋅2⋅1) = 56.

56⋅8 = 448

So the fourth term is -448a5b3.