prof by binomial theorem

Question

 

by using binomial theorem prof that

C (100،0) + C (100،2) + C (100،4) + ... + C (100،100) = C (100،1) + C (100،3) + C (100،5) + ... + C (100،99)

Answer 1

(a-b)^n=a^n-nba^(n-1)+n(n-1)b^2a^(n-2)-... which can also be written a^n*nC0-ba^(n-1)*nC1+b^2a^(n-2)*nC2-...,
where the mathematical symbol for combination of r distinct items out of n, nCr=C (n،r).

If a=b=1 and n=100 we have (1-1)^100=0=100C0-100C1+100C2-100C3+...-100C99+100C100.

Therefore, taking all the negative terms to the other side of the equation, 100C1+100C3+100C5+...100C99=100C0+100C2+100C4+...100C100 QED.