Prove that sum $\sum_{k=1}^n {(-1)^k {{n-1} \choose {k-1}} (2n-k-1) 2^k}$ is zero I try to prove that $\sum_{k=1}^n {(-1)^k {{n-1} \choose {k-1}} (2n-k-1) 2^k}=0.$
I calculated in Maple for n=1..100.
 $\sum_{k=1}^n {(-1)^k {{n-1} \choose {k-1}} (2n-k-1) 2^k}=-2  \sum_{k=0}^{n-1} {(-1)^k {{n-1} \choose {k}} (2n-k-2) 2^k}.$
I am not sure how to continue
I assume that this has something to do with the binomial theorem.
Can anyone help me with this? 
 A: Hint:
$$
\begin{align*}
&\sum_{k=1}^{n} (-2)^{k-1} \binom{n-1}{k-1} (n-k) = (n-1) \sum_{k=1}^{n-1} (-2)^k \binom{n-2}{k-1} = (-1)^{n-2} (n-1), \\
&\sum_{k=1}^{n} (-2)^{k-1} \binom{n-1}{k-1} = (-1)^{n-1}.
\end{align*}
$$
Therefore
$$
\sum_{k=1}^n (-2)^{k-1} \binom{n-1}{k-1} (n-k+n-1) = (-1)^{n-2} (n-1) + (-1)^{n-1} (n-1) = 0.
$$
A: I'll name $S_n$ your sum. Let's change the index: $k \leftarrow k-1$
$$ S_n = -2\sum_{k=0}^{n-1} {(-1)^{k} {{n-1} \choose {k}} (2*(n-1)-k) 2^{k}} $$
$$ S_n = -2[2(n-1)*(-1)^{n-1}\sum_{k=0}^{n-1} {(-1)^{n-1-k} {{n-1} \choose {k}} 2^{k}} +\sum_{k=0}^{n-1} {(-1)^{k} {{n-1} \choose {k}} k*2^{k}}] $$
$1 = (2-1)^{n-1} = \sum_{k=0}^{n-1} {(-1)^{n-1-k} {{n-1} \choose {k}} 2^{k}}$
$$=> S_n = -2[2(n-1)*(-1)^{n-1} +2(n-1)\sum_{k=1}^{n-1} {(-1)^{k} {{n-2} \choose {k-1}} *2^{k-1}}] $$
$$ S_n = -2[2(n-1)*(-1)^{n-1} -2(n-1)(-1)^{n-1}\sum_{k=0}^{n-2} {(-1)^{n-2-k} {{n-2} \choose {k}} *2^{k}}] $$
$1 = (2-1)^{n-2} = \sum_{k=0}^{n-2} {(-1)^{n-2-k} {{n-2} \choose {k}} *2^{k}} $
$$=> S_n = -2[2(n-1)*(-1)^{n-1} -2(n-1)(-1)^{n-1}] = 0 $$
