<s>[INST] 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!
 [/INST] 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
$$\begin{align}
2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1} &= \sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k}\\
&= \left(\left( \sum_{k=0}^{n}(-1)^k\binom{n-k-1}{k}2^{n-2k} \right) - \left( -1^{n}\binom{n-n-1}{n}2^{n-2n}\right) \right)\\
&= \frac{2^{n+1}\cdot n + 1}{2^n} - \frac{ -1^n \binom{-1}{n}}{2^n}\\
&= \frac{2^{n+1}\cdot n + 1 + (-1^{n-1}) \binom{-1}{n} }{2^n}\\
\end{align}$$
and
$$\begin{align}
\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2} &= 2^{-1} \sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-1}\\
&=2^{-1} \left( \left( \sum_{k=0}^{n-1} (-1)^k\binom{n-k-2}{k}2^{n-2k-1} \right) - \left(  (-1)^{n-1}\binom{n-(n-1)-2}{n-1}2^{n-2(n-1)-1}  \right)\right)\\
&= 2^{-1} \left( \left( \frac{2^{(n-1)+1}\cdot (n-1) + 1}{2^{n-1}} \right) -\left( -1^{n-1} \binom{-1}{n-1}2^{-n+1} \right) \right)\\
&= \frac{2^n\cdot (n-1) + 1}{2^n} -\left( -1^{n-1} \binom{-1}{n-1}2^{-n} \right)\\
&= \frac{2^n\cdot (n-1) + 1}{2^n} - \frac{-1^{n-1} \binom{-1}{n-1}}{2^n}\\
&= \frac{2^n\cdot (n-1) + 1 +(-1^n)\binom{-1}{n-1}}{2^n}
\end{align}$$
thus you can rewrite the RHS of your statement as
$$\begin{align}
RHS &= 2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2} \\
&= \frac{2^{n+1}\cdot n + 1 + (-1^{n-1}) \binom{-1}{n} }{2^n} - \frac{2^n\cdot (n-1) + 1 +(-1^n) \binom{-1}{n-1}}{2^n}\\
&=  \frac{2^{n+1}\cdot n + 1 + (-1^{n-1}) \binom{-1}{n} - 2^n\cdot (n-1) - 1 - (-1^n) \binom{-1}{n-1}}{2^n}\\
&=  \frac{2^n(n+1) + (-1^{n-1}) \binom{-1}{n} + (-1^{n-1}) \binom{-1}{n-1}}{2^n} 
\end{align}$$
As for the LHS you should easily be able to rewrite it, and check if the statement is true.
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