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)(am+bm+k=1m-1(mk)am-kbk) by the inductive hypothesis
= am+1+bm+1+abm+bam+k=1m-1(mk)am-k+1bk+k=1m-1(mk)am-kbk+1
= am+1+bm+1+k=1m(mk)am-k+1bk+k=0m-1(mk)am-kbk+1 by combining terms
= am+1+bm+1+k=1m(mk)am-k+1bk+j=1m(mj-1)am+1-jbj let j=k+1 in second sum
= am+1+bm+1+k=1m[(mk)+(mk-1)]am+1-kbk by combining the sums
= am+1+bm+1+k=1m(m+1k)am+1-kbk 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