inductive proof of binomial theorem

We prove the theorem for a ring. We do not assume a unit for the ring. We do not need commutativity of the ring, but only that $a$ and $b$ commute.

When $n=1$, the result is clear.

For the inductive step, assume it holds for $m$. Then for $n=m+1$,

${(a+b)}^{m+1}$ $=$ $(a+b){(a+b)}^m$

$=$ $(a+b)(a^m+b^m+{\displaystyle \sum _{k=1}^{m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k}{b}^k)\text{ by the inductive hypothesis}$

$=$ $a^{m+1}+b^{m+1}+a{b}^m+b{a}^m+{\displaystyle \sum _{k=1}^{m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k+1}{b}^k+{\displaystyle \sum _{k=1}^{m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k}{b}^{k+1}$

$=$ $a^{m+1}+b^{m+1}+{\displaystyle \sum _{k=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k+1}{b}^k+{\displaystyle \sum _{k=0}^{m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k}{b}^{k+1}\text{ by combining terms}$

$=$ $a^{m+1}+b^{m+1}+{\displaystyle \sum _{k=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k+1}{b}^k+{\displaystyle \sum _{j=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j-1}}\right)a^{m+1-j}{b}^j\text{ let j=k+1 in second sum}$

$=$ $a^{m+1}+b^{m+1}+{\displaystyle \sum _{k=1}^m}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k-1}}\right)\right]a^{m+1-k}{b}^k\text{ by combining the sums}$

$=$ $a^{m+1}+b^{m+1}+{\displaystyle \sum _{k=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m+1}{k}}\right)a^{m+1-k}{b}^k\text{ from Pascal’s rule}$

as desired.

Title	inductive proof of binomial theorem
Canonical name	InductiveProofOfBinomialTheorem
Date of creation	2013-03-22 11:48:06
Last modified on	2013-03-22 11:48:06
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	21
Author	Mathprof (13753)
Entry type	Proof
Classification	msc 05A10