SOLUTION: Show that nC0+nC2+nC4+...=nC1+nC3+...=2^,where ^=n-1.

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Question 97167: Show that nC0+nC2+nC4+...=nC1+nC3+...=2^,where ^=n-1.
Answer by mathslover(157) About Me  (Show Source):
You can put this solution on YOUR website!
we have
+%28+1+%2B+x+%29%5E+n = nC0 + nC1x + nC2x^2 + nC3x^3 + ...)
Sustituting x= -1
we have 0%5En = 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%5E%28n-1%29
Substitute x =1 in the expansion of +%28+1+%2B+x+%29%5E+n
we have %282%5En%29 = nC0 + nC1 + nC2 + nC3 + .......
%282%5En%29 = (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
%282%5En%29 = (nC0 + nC2 + nC4 +...) + (nC0 + nC2 + nC4 +...)
2(nC0 + nC2 + nC4 +...) = %282%5En%29
Dividing by 2 on both sides
(nC0 + nC2 + nC4 +...) = %282%5En%29%2F2 = 2%5E%28n-1%29
therefore we have
(nC0 + nC2 + nC4 +...) = (nC1 + nC3 + nC5 +.... ) = 2%5E%28n-1%29