Find a closed expression for $\displaystyle\sum_{k=\frac {n+2} 2} ^n \binom n k$, $n$ is even.
My attempt:
$(1+1)^n = \displaystyle\sum_{k=0} ^ n \binom n k= \sum_{k=0} ^{\frac {n-2} 2}\binom n k + \binom n {\frac n 2} + \sum_{k=\frac {n+2} 2} ^n\binom n k$
From the identity $\binom n k = \binom n {n-k}$ we know that: $A=\displaystyle\sum_{k=0} ^{\frac {n-2} 2}\binom n k = \sum_{k=\frac {n+2} 2} ^n\binom n k$
So we have: $2^n = 2A + \binom n {\frac n 2} \Rightarrow A = 2^{n-1} - \frac 1 2 \binom n {\frac n 2}$
But I was told that something is missing in this approach, but I can't see what can be wrong here.