Equality of two binomial coefficient containing expressions 
Why is 
$$ \begin{align}
 &\sum_{k=0}^n(-1)^k\left[\binom{n-k-1}{k}+\binom{n-k-1}{k-1}\right]2^{n-2k}\\
 &=2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2}\\[6pt] & \end{align} $$ ?

Should'nt it be $$\begin{align}\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k}-\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k-1}2^{n-2k}\\[6pt] & \end{align} $$
I know this must be a silly question but I can't understand this...please help me out!

First,
$$\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n-k-1}k2^{n-2k}&=\sum_{k=0}^n(-1)^k\binom{n-k-1}k2\cdot2^{n-2k-1}\\
&=2\sum_{k=0}^n(-1)^k\binom{n-k-1}k2^{n-2k-1}\\
&=2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}k2^{n-2k-1}\;,
\end{align*}$$
since $\binom{n-k-1}k=0$ when $k=n$. Then, setting $\ell=k-1$, so that $k=\ell+1$, we have
$$\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n-k-1}{k-1}2^{n-2k}&=\sum_{\ell=-1}^{n-1}(-1)^{\ell+1}\binom{n-(\ell+1)-1}\ell2^{n-2(\ell+1)}\\
&=(-1)\sum_{\ell=-1}^{n-1}(-1)^\ell\binom{n-\ell-2}\ell2^{n-2\ell-2}\\
&=-\sum_{\ell=0}^{n-2}(-1)^\ell\binom{n-\ell-2}\ell2^{n-2\ell-2}\;,\tag{1}
\end{align*}$$
since $\binom{n-\ell-2}\ell=0$ when $\ell=-1$ and when $\ell=n-1$. Now just rename $\ell$ to $k$ in $(1)$, and you have the desired result:
$$\begin{align*}
&\sum_{k=0}^n(-1)^k\left[\binom{n-k-1}k+\binom{n-k-1}{k-1}\right]2^{n-2k}\\
&\qquad=2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}k2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}k2^{n-2k-2}\;.
\end{align*}$$
