Thread: [SOLVED] combination prove ques

( The better form of the question with correct symbols is here.
http://www.geocities.com/johntyung/Q2.pdf )


*** For this problem I heard that they solved it by just using the identity and manipulating it algebraically. Without using induction. ***

for n even, prove that

(n choose 0) + (n choose 2) + (n choose 4) +...+ (n choose n)=2^(n-1)

and

(n choose 1) + (n choose 3) +...+(n choose (n-1)) = 2^(n-1)

using mathematical induction and/or various identities involving binomial coefficients.
(Hint: first establish the identity,

((m+2) choose k) = (m choose k)+ 2(m choose (k-1)) +(m choose (k-2) )

Then, when doing the induction, do a simultaneous induction on both identities.

This is too complicated. Prove that

nC0 + nC1 + nC2 + nC3 + ... + nCn = 2^n

and

nC0 - nC1 + nC2 - nC3 + ... + nCn = 0 (assuming n is even).

Then add and subtract these two identities. They follow from 1+1 = 2 and 1-1 = 0 and the Binomial Theorem.

Looks like a binomial series with a and b equal to 1, http://tutorial.math.lamar.edu/AllBr...mialSeries.asp