The Binomial Theorem is easier and more naturally proven in a combinatorics context but can be proven by induction.

Problem: Prove the Binomial Theorem by Induction.

Solution: Let P(n) be the proposition that for x,y\in\mathbb{R}, n\in\mathbb{N}

(x+y)^n=\sum_{i=0}^n{n\choose i}x^{n-i}y^i

(Binomial Theorem)


\sum_{i=0}^1{1\choose i}x^{1-i}y^i={1\choose 0}x^{1-0}y^0=x+y=(x+y)^1

(P(1) is true)

Now assume P(k) is true; that is:

(x+y)^k=\sum_{i=0}^k{k\choose i}x^{k-i}y^i



=\left(\sum_{i=0}^k{k\choose i}x^{k-i}y^i\right)(x+y)

=\underbrace{\left(\sum_{i=0}^k{k\choose i}x^{k+1-i}y^i\right)}_{=S_1}+\underbrace{\left(\sum_{i=0}^k{k\choose i}x^{k-i}y^{i+1}\right)}_{=S_2}

Now all terms are of the form c(i)x^{k+1-i}y^i as i runs from 0\rightarrow k+1. Let j\in\{0,1,\dots,k+1\}. Now the x^{k+1-j}y^j term has constant from S_1 and S_2:

x^{k+1-j}y^j\left[{k\choose j}+{k\choose j-1}\right]

\Rightarrow (x+y)^{k+1}=\sum_{i=0}^{k+1}\left[{k\choose j}+{k\choose j-1}\right]x^{k+1-j}y^j

It is a straightforward exercise to show:

{n+1\choose k}={n\choose k}+{n\choose k-1}


(x+y)^{k+1}=S_1+S_2=\sum_{i=0}^{k+1}{k+1\choose i}x^{k+1-i}y^ii

(P(k+1) is true)

P(1) is true. P(k)\Rightarrow P(k+1). Hence P(n) is true for all n\in\mathbb{N}; i.e. the Binomial Theorem is true \bullet