Overview

The Binomial Theorem is one of the most important bridges between algebra and combinatorics.

It gives a formula for expanding expressions of the form

(a+b)n.(a+b)^n.

For small powers, we can expand directly:

(a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2

and

(a+b)3=a3+3a2b+3ab2+b3.(a+b)^3=a^3+3a^2b+3ab^2+b^3.

The Binomial Theorem generalises this pattern to any nonnegative integer power nn.

It explains:

  • Why the coefficients in these expansions appear
  • How to find a specific term without expanding everything
  • How powers can be estimated or simplified
  • How algebraic expansions connect to counting and probability

Motivation

Suppose we want to expand

(x+2)8.(x+2)^8.

Multiplying out eight copies by hand would be slow:

(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2).(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2).

However, the Binomial Theorem lets us write the full expansion immediately.

It tells us that each term comes from choosing some copies to contribute xx and the remaining copies to contribute 22.

For example, the term involving x5x^5 occurs when we choose xx from 55 of the factors and 22 from the other 33 factors. The number of ways to choose which 55 factors provide xx is

(85).\binom{8}{5}.

So the x5x^5 term is

(85)x523.\binom{8}{5}x^5 2^3.

This is the main idea of the theorem, expansion coefficients count choices.

Binomial Coefficients

Before stating the theorem, we recall the notation

(nk).\binom{n}{k}.

This is read as nn choose kk and is defined by

(nk)=n!k!(n−k)!.\binom{n}{k} = \frac{n!}{k!(n-k)!}.

It counts the number of ways to choose kk objects from a set of nn objects.

For example,

(52)=5!2!3!=10.\binom{5}{2} = \frac{5!}{2!3!} = 10.

So there are 1010 ways to choose 22 objects from 55.

Useful special values include

(n0)=1\binom{n}{0}=1

and

(nn)=1.\binom{n}{n}=1.

Also,

(nk)=(nn−k).\binom{n}{k}=\binom{n}{n-k}.

This symmetry comes from the fact that choosing kk objects to include is the same as choosing n−kn-k objects to exclude.

The Binomial Theorem

For every nonnegative integer nn,

(a+b)n=∑k=0n(nk)an−kbk.(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k.

Written out, this means

(a+b)n=(n0)an+(n1)an−1b+(n2)an−2b2+⋯+(nn)bn.(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n.

The general term is

(nk)an−kbk.\binom{n}{k}a^{n-k}b^k.

Here, kk tells us how many times we choose bb from the nn factors.

Why the Formula Works

Consider

(a+b)n.(a+b)^n.

This means

(a+b)(a+b)⋯(a+b),(a+b)(a+b)\cdots(a+b),

with nn factors.

To form a term in the expansion, we choose either aa or bb from each factor.

Suppose we want a term containing exactly kk copies of bb. Then we must choose bb from kk of the nn factors and aa from the remaining n−kn-k factors.

This produces

an−kbk.a^{n-k}b^k.

The number of ways to choose which kk factors supply bb is

(nk).\binom{n}{k}.

Therefore, the coefficient of

an−kbka^{n-k}b^k

is

(nk).\binom{n}{k}.

This proves the Binomial Theorem.

Pascal's Triangle

The coefficients in binomial expansions form Pascal's Triangle:

11
111\quad 1
1211\quad 2\quad 1
13311\quad 3\quad 3\quad 1
146411\quad 4\quad 6\quad 4\quad 1
151010511\quad 5\quad 10\quad 10\quad 5\quad 1

The row beginning with 1,4,6,4,11,4,6,4,1 gives the coefficients of

(a+b)4.(a+b)^4.

So

(a+b)4=a4+4a3b+6a2b2+4ab3+b4.(a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4.

Each entry in Pascal's Triangle is the sum of the two entries above it. This reflects the identity

(nk)=(n−1k−1)+(n−1k).\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}.

This identity is often useful in combinatorial arguments.

Expanding Powers

The most direct use of the Binomial Theorem is expanding powers.

Example 1

Expand (x+3)5(x+3)^5.

Using the Binomial Theorem,

(x+3)5=∑k=05(5k)x5−k3k.(x+3)^5 = \sum_{k=0}^{5} \binom{5}{k}x^{5-k}3^k.

The coefficients in row 55 of Pascal's Triangle are

1,5,10,10,5,1.1,5,10,10,5,1.

Therefore,

(x+3)5=x5+15x4+90x3+270x2+405x+243.(x+3)^5 = x^5+15x^4+90x^3+270x^2+405x+243.

Example 2

Expand (2x−y)4(2x-y)^4.

Here, take

a=2xa=2x

and

b=−y.b=-y.

Then

(2x−y)4=∑k=04(4k)(2x)4−k(−y)k.(2x-y)^4 = \sum_{k=0}^{4} \binom{4}{k}(2x)^{4-k}(-y)^k.

Now compute each term:

(2x)4=16x4(2x)^4=16x^4
4(2x)3(−y)=−32x3y4(2x)^3(-y)=-32x^3y
6(2x)2(−y)2=24x2y26(2x)^2(-y)^2=24x^2y^2
4(2x)(−y)3=−8xy34(2x)(-y)^3=-8xy^3
(−y)4=y4.(-y)^4=y^4.

As such

(2x−y)4=16x4−32x3y+24x2y2−8xy3+y4.(2x-y)^4 = 16x^4-32x^3y+24x^2y^2-8xy^3+y^4.

Finding a Specific Term

Often we do not need the entire expansion. We only need one coefficient or one term.

The general term in

(a+b)n(a+b)^n

is

(nk)an−kbk.\binom{n}{k}a^{n-k}b^k.

This is the term containing bkb^k.

Example 3

Find the coefficient of x7x^7 in (x+2)10(x+2)^{10}.

The general term is

(10k)x10−k2k.\binom{10}{k}x^{10-k}2^k.

We want the exponent of xx to be 77, so

10−k=7.10-k=7.

Thus

k=3.k=3.

The coefficient is

(103)23.\binom{10}{3}2^3.

Now

(103)=120,\binom{10}{3}=120,

so the coefficient is

120(8)=960.120(8)=960.

Example 4

Find the constant term in

(x+2x)6.\left(x+\frac{2}{x}\right)^6.

The general term is

(6k)x6−k(2x)k.\binom{6}{k}x^{6-k}\left(\frac{2}{x}\right)^k.

This becomes

(6k)2kx6−k−k=(6k)2kx6−2k.\binom{6}{k}2^k x^{6-k-k} = \binom{6}{k}2^k x^{6-2k}.

For the constant term, we need the exponent of xx to be 00:

6−2k=0.6-2k=0.

Thus

k=3.k=3.

The constant term is

(63)23=20(8)=160.\binom{6}{3}2^3=20(8)=160.

Alternating Signs

The Binomial Theorem also handles expressions of the form

(a−b)n.(a-b)^n.

Since

a−b=a+(−b),a-b=a+(-b),

we get

(a−b)n=∑k=0n(nk)an−k(−b)k.(a-b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}(-b)^k.

So the signs alternate depending on the parity of kk.

For example,

(a−b)5=a5−5a4b+10a3b2−10a2b3+5ab4−b5.(a-b)^5 = a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5.

A common mistake is to get these signs wrong, especially in middle terms.

Part 2

Read the continuation in Part 2!

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