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 $n=m+1$,

 $\displaystyle(a+b)^{m+1}$ $\displaystyle=$ $\displaystyle(a+b)(a+b)^{m}$ $\displaystyle=$ $\displaystyle(a+b)(a^{m}+b^{m}+\sum_{k=1}^{m-1}\binom{m}{k}a^{m-k}b^{k})\text{% by the inductive hypothesis}$ $\displaystyle=$ $\displaystyle a^{m+1}+b^{m+1}+ab^{m}+ba^{m}+\sum_{k=1}^{m-1}\binom{m}{k}a^{m-k% +1}b^{k}+\sum_{k=1}^{m-1}\binom{m}{k}a^{m-k}b^{k+1}$ $\displaystyle=$ $\displaystyle a^{m+1}+b^{m+1}+\sum_{k=1}^{m}\binom{m}{k}a^{m-k+1}b^{k}+\sum_{k% =0}^{m-1}\binom{m}{k}a^{m-k}b^{k+1}\text{ by combining terms}$ $\displaystyle=$ $\displaystyle a^{m+1}+b^{m+1}+\sum_{k=1}^{m}\binom{m}{k}a^{m-k+1}b^{k}+\sum_{j% =1}^{m}\binom{m}{j-1}a^{m+1-j}b^{j}\text{ let j=k+1 in second sum}$ $\displaystyle=$ $\displaystyle a^{m+1}+b^{m+1}+\sum_{k=1}^{m}\left[\binom{m}{k}+\binom{m}{k-1}% \right]a^{m+1-k}b^{k}\text{ by combining the sums}$ $\displaystyle=$ $\displaystyle a^{m+1}+b^{m+1}+\sum_{k=1}^{m}\binom{m+1}{k}a^{m+1-k}b^{k}\text{% from Pascal's rule}$

as desired.

Title inductive proof of binomial theorem InductiveProofOfBinomialTheorem 2013-03-22 11:48:06 2013-03-22 11:48:06 Mathprof (13753) Mathprof (13753) 21 Mathprof (13753) Proof msc 05A10