Binomial and combinatorics

Question

Please help me solve this problem.

Show that:

$$\sum_{k=1}^n k{n \choose 2k+1} = (n-2)2^{n-3}$$

Answer 1

$$\frac{(1+x)^n-(1-x)^n}{2}=\sum_{k=0}^{n} {n \choose 2k+1} x^{2k+1}$$ $$\frac{(1+x)^n-(1-x)^n}{2x}=\sum_{k=0}^{n} {n \choose 2k+1} x^{2k}$$ Differentiate w.r.t. $x$ both sides $$\frac{n(1+x)^{n-1}+n(1-x)^{n-1}}{2x}+\frac{(1+x)^n-(1-x)^n}{-2x^2}=\sum_{k=0}^{n} 2k {n \choose 2k+1} x^{2k-1}$$ Now put $x=1$ in above, to get $$\sum_{k=0}^{n} k{n \choose 2k+1}=2^{n-3}(n-2)$$

the upper limit of the sum could be $[n/2]$ or $n$.

Answer 2

Consider $[n]= \{ 1,2, \cdots,n\}$ with their usual order.

Choose an element of $[n]$ excluding the first and last elements. Now choose an odd number of elements before and an odd number of elements after. It easy to see that this enumerates to $(n-2)2^{n-3}$.

Alternatively choose an odd ($ \geq 3$) number of elements from $[n]$ and now choose on of the "even" elements from this subset.