Proving $\sum_{j=0}^n (-1)^j {n \choose j} F_{s+2n-2j} = F_{s+n} $, where $F_n$ is the $n$-th Fibonacci number 
$$\sum_{j=0}^n (-1)^j {n \choose j} F_{s+2n-2j} = F_{s+n} $$
($F$ is Fibonacci number).

I have been trying to  prove this by mathematical induction.
First I assume this is true for n.
If I could prove it works for n+1, then it is done.
$\sum_{j=0}^{n+1} (-1)^j {n+1 \choose j} F_{s+2n-2j+2} = F_{s+n+1} $
Also I could replace s with s+1 since it is valid for general constant s.
So I could get $\sum_{j=0}^n (-1)^j {n \choose j} F_{s+1+2n-2j} = F_{s+1+n} $

Here is a proof by induction over $n$.  To recap, we want to show
$$F_{s+n} = \sum_{j=0}^n (-1)^j \binom{n}{j} F_{s+2n-2j} \tag{1}$$
The base case, $n=0$, is trivial.  So suppose (1) holds for some value of $n$.  We want show it holds for $n+1$.  Since $s$ is arbitrary, we may substitute $s+1$ for $s$ in (1):
$$F_{s+n+1} = \sum_{j=0}^n (-1)^j \binom{n}{j} F_{s+1+2n-2j}$$
Next, we apply the Fibonacci identity $F_{n} = F_{n+1}-F_{n-1}$:
$$F_{s+n+1} = \sum_{j=0}^n (-1)^j \binom{n}{j} (F_{s+2+2n-2j}-F_{s+2n-2j})$$
So
$$\begin{align}
F_{s+n+1} &= \sum_{j=0}^n (-1)^j \binom{n}{j} F_{s+2+2n-2j} - \sum_{j=0}^n (-1)^j \binom{n}{j} F_{s+2n-2j} \\
&= F_{s+2+2n} +  \sum_{j=1}^n (-1)^j \binom{n}{j} F_{s+2+2n-2j} - \sum_{j=0}^{n-1} (-1)^j \binom{n}{j} F_{s+2n-2j} - (-1)^n F_s \\
&= F_{s+2+2n} +  \sum_{j=1}^n (-1)^j \binom{n}{j} F_{s+2+2n-2j} - \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{j-1} F_{s+2+2n-2j} - (-1)^n F_s \tag{2}\\
&= F_{s+2+2n} +  \sum_{j=1}^n (-1)^j \left[ \binom{n}{j} + \binom{n}{j-1} \right] F_{s+2+2n-2j} +(-1)^{n+1}F_s \\
&= F_{s+2+2n} +  \sum_{j=1}^n (-1)^j  \binom{n+1}{j}  F_{s+2+2n-2j} +(-1)^{n+1}F_s \tag{3}\\
&= \sum_{j=0}^{n+1} (-1)^j \binom{n+1}{j} F_{s+2(n+1)-2j}
\end{align}$$
This completes the proof by induction.
Notes 
(2) Shifting the index in the second sum 
(3) Applying the recursive formula for binomial coefficients
