Evaluate the sum ${n \choose 1} + 3{n \choose 3} +5{n \choose 5} + 7{n \choose 7}...$ in closed form How do I evaluate the sum:$${n \choose 1} + 3{n \choose 3} +5{n \choose 5} + 7{n \choose 7}  ...$$in closed form?
I don't really know how to start and approach this question. Any help is greatly appreciated.
 A: Starting with 
\begin{align}
(1+t)^{n} = \sum_{k=0}^{n} \binom{n}{k} t^{k}
\end{align}
it is seen that
\begin{align}
\frac{1}{2} \left[ (1+t)^{n} - (1-t)^{n} \right] = \sum_{k=1}^{[(n+1)/2]} \binom{n}{2k-1} t^{2k-1}.
\end{align}
Now differentiating both sides leads to
\begin{align}
\sum_{k=1}^{[(n+1)/2]} (2k-1) \binom{n}{2k-1} t^{2k-2} = \frac{n}{2} \left[ (1+t)^{n-1} + (1-t)^{n-1} \right]
\end{align}
and upon setting $t=1$ the desired result is
\begin{align}
\sum_{k=1}^{[(n+1)/2]} (2k-1) \binom{n}{2k-1} = 2^{n-2} \ n.
\end{align}  
A: Hint: We have $\binom{n}{2k-1}=\frac{n}{2k-1}\binom{n-1}{2k-2}$.
Note that $\binom{a}{b}=0$ if $b\gt a$.
A: Here is a combinatorial proof because I like combinatorial proofs.
Note that that $\sum_{k \geq 0} \binom{n}{2k} = 2^{n-1}$. One way to prove this result is with the binomial theorem. We will use this fact.
Now $\sum_{k \geq 0} (2k+1)\binom{n}{2k+1}$ can be interpreted as choosing a subset of odd size from a set with $n$ elements in one of $\binom{n}{2k+1}$ ways and then choosing a distinguished element from such a subset in one of $2k+1$ ways. 
Another way to make the same choice would be to first choose the distinguished element in one of $n$ ways and then choose the remaining elements of such a subset in one of $\binom{n-1}{2k}$ ways. Thus we have
$$\sum_{k \geq 0} (2k+1)\binom{n}{2k+1} =  n \sum_{k\geq 0} \binom{n-1}{2k} = n 2^{n-2}.$$
Notice this solution essentially uses the hint Did gave in the comments. The identity Did provides can be proven combinatorially as we have or algebraically.
