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


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

$$\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n-k-1}k2^{n-2k}&amp;=\sum_{k=0}^n(-1)^k\binom{n-k-1}k2\cdot2^{n-2k-1}\\
&amp;=2\sum_{k=0}^n(-1)^k\binom{n-k-1}k2^{n-2k-1}\\
&amp;=2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}k2^{n-2k-1}\;,
\end{align*}$$

since $\binom{n-k-1}k=0$ when $k=n$. Then, setting $\ell=k-1$, so that $k=\ell+1$, we have

$$\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n-k-1}{k-1}2^{n-2k}&amp;=\sum_{\ell=-1}^{n-1}(-1)^{\ell+1}\binom{n-(\ell+1)-1}\ell2^{n-2(\ell+1)}\\
&amp;=(-1)\sum_{\ell=-1}^{n-1}(-1)^\ell\binom{n-\ell-2}\ell2^{n-2\ell-2}\\
&amp;=-\sum_{\ell=0}^{n-2}(-1)^\ell\binom{n-\ell-2}\ell2^{n-2\ell-2}\;,\tag{1}
\end{align*}$$

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:

$$\begin{align*}
&amp;\sum_{k=0}^n(-1)^k\left[\binom{n-k-1}k+\binom{n-k-1}{k-1}\right]2^{n-2k}\\
&amp;\qquad=2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}k2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}k2^{n-2k-2}\;.
\end{align*}$$
This is a partial answer

Since
$$\sum_{k=0}^n(-1)^k \binom{n-k-1}{k} 2^{n-2k} = \frac{2^{n+1}\cdot n + 1}{2^n}$$
you get immediately that
$$\begin{align}
2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1} &amp;= \sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k}\\
&amp;= \left(\left( \sum_{k=0}^{n}(-1)^k\binom{n-k-1}{k}2^{n-2k} \right) - \left( -1^{n}\binom{n-n-1}{n}2^{n-2n}\right) \right)\\
&amp;= \frac{2^{n+1}\cdot n + 1}{2^n} - \frac{ -1^n \binom{-1}{n}}{2^n}\\
&amp;= \frac{2^{n+1}\cdot n + 1 + (-1^{n-1}) \binom{-1}{n} }{2^n}\\
\end{align}$$
and
$$\begin{align}
\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2} &amp;= 2^{-1} \sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-1}\\
&amp;=2^{-1} \left( \left( \sum_{k=0}^{n-1} (-1)^k\binom{n-k-2}{k}2^{n-2k-1} \right) - \left(  (-1)^{n-1}\binom{n-(n-1)-2}{n-1}2^{n-2(n-1)-1}  \right)\right)\\
&amp;= 2^{-1} \left( \left( \frac{2^{(n-1)+1}\cdot (n-1) + 1}{2^{n-1}} \right) -\left( -1^{n-1} \binom{-1}{n-1}2^{-n+1} \right) \right)\\
&amp;= \frac{2^n\cdot (n-1) + 1}{2^n} -\left( -1^{n-1} \binom{-1}{n-1}2^{-n} \right)\\
&amp;= \frac{2^n\cdot (n-1) + 1}{2^n} - \frac{-1^{n-1} \binom{-1}{n-1}}{2^n}\\
&amp;= \frac{2^n\cdot (n-1) + 1 +(-1^n)\binom{-1}{n-1}}{2^n}
\end{align}$$
thus you can rewrite the RHS of your statement as
$$\begin{align}
RHS &amp;= 2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2} \\
&amp;= \frac{2^{n+1}\cdot n + 1 + (-1^{n-1}) \binom{-1}{n} }{2^n} - \frac{2^n\cdot (n-1) + 1 +(-1^n) \binom{-1}{n-1}}{2^n}\\
&amp;=  \frac{2^{n+1}\cdot n + 1 + (-1^{n-1}) \binom{-1}{n} - 2^n\cdot (n-1) - 1 - (-1^n) \binom{-1}{n-1}}{2^n}\\
&amp;=  \frac{2^n(n+1) + (-1^{n-1}) \binom{-1}{n} + (-1^{n-1}) \binom{-1}{n-1}}{2^n} 
\end{align}$$
As for the LHS you should easily be able to rewrite it, and check if the statement is true.