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We know that $\sum_{i=0}^{n} {{n}\choose{i}} =2^n$  We also know that the alternation sum of binomial coefficients from (0,n)=0 (which is summation (i=0,n) of (-1)^i $\sum_{i=0}^{n} -1^i  * nCi) {{n}\choose{i}} =0$  If you subtract the second equation from the first you get   2*summation from (i=0,n) of nC(2i+1)=2^n  which implies that summation from (i=0,n) of nC(2i+1)=2^(n-1)