Question 97167
 we have 

{{{ ( 1 + x )^ n}}} = nC0 + nC1x + nC2x^2 + nC3x^3 + ...)

Sustituting x= -1 

we have {{{0^n}}} = nC0 - nC1 + nC2 - nC3 + ...

0 =  nC0 - nC1 + nC2 - nC3 + ...

grouping all the negative terms on the other side we have 

nC0 + nC2 + nC4 + .... = nC1 + nC3 + nC5 +.... which completes the first portion of the proof .

to prove that each of these expressions evaluate to {{{2^(n-1)}}}

Substitute x =1 in the expansion of {{{ ( 1 + x )^ n}}} 

we have {{{(2^n)}}} = nC0 + nC1 + nC2 + nC3 + .......

{{{(2^n)}}} = (nC0 + nC2 + nC4 +...) + (nC1 + nC3 + nC5 +...) grouping the odd and even terms together

Since (nC0 + nC2 + nC4 +...) = (nC1 + nC3 + nC5 +.... ) as proved already we can rewrite the expression above as 

{{{(2^n)}}} = (nC0 + nC2 + nC4 +...) + (nC0 + nC2 + nC4 +...)

2(nC0 + nC2 + nC4 +...) = {{{(2^n)}}} 
Dividing by 2 on both sides
(nC0 + nC2 + nC4 +...) = {{{(2^n)/2}}} = {{{2^(n-1)}}} 

therefore we have 
(nC0 + nC2 + nC4 +...) = (nC1 + nC3 + nC5 +.... ) = {{{2^(n-1)}}}