Does $\sum_{k=0}^{2n}(-1)^k{2n\choose k}2^kF_{k+1}=5^n$ hold for all n values? I was looking at formula $(81)$ on here which is shown below
$$\sum_{k=0}^{n}{2n\choose k}2^kF_k=F_{3n}$$.
$F_n$ is the $n^{th}$ Fibonacci number
I wrote out the sum and just alternate the signs and found out that it has a simple answer in the form of $5^n$
$$\sum_{k=0}^{2n}(-1)^k{2n\choose k}2^kF_{k+1}=5^n$$.
Here notice that it is only work for even terms.
Does this formula $$\sum_{k=0}^{2n}(-1)^k{2n\choose k}2^kF_{k+1}=5^n$$
works for all n values or it just accidentally for some n values?
Examples

for $n=1,2$ and $3$
$$1-4+8=5$$
$$1-8+48-96+80=5^2$$
$$1-12+120-480+1200-1536+832=5^3$$
 A: Yes, the identity holds and is easily proven with induction on $n$.  We first recall the familiar identity $$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1},$$ which is of course simply Pascal's triangle.  If we apply this twice, we get $$\binom{n}{k-1} + 2\binom{n}{k} + \binom{n}{k+1} = \binom{n+2}{k+1},$$ which is equivalent to $$\binom{2n}{k-1} + 2\binom{2n}{k} + \binom{2n}{k+1} = \binom{2(n+1)}{k+1}.$$  Now define $$S(n) = \sum_{k=0}^{2n} (-2)^k \binom{2n}{k} F_{k+1}.$$  Then recalling the convention that $\binom{n}{k} = 0$ when $k < 0$ or $k > n$, 
$$\begin{align*} S(n+1) &= \sum_{k=0}^{2(n+1)} (-2)^k \binom{2(n+1)}{k} F_{k+1} \\ 
&= \sum_{k=0}^{2(n+1)} (-2)^k \binom{2n}{k-2}F_{k+1} + 2 \sum_{k=0}^{2(n+1)} (-2)^k \binom{2n}{k-1}F_{k+1} + S(n) \\
&= S(n) + 4\sum_{k=0}^{2(n+1)} (-2)^{k-2} \binom{2n}{k-2} (2F_{k-1} + F_{k-2}) - 4 \sum_{k=0}^{2(n+1)} (-2)^{k-1} \binom{2n}{k-1} (F_k + F_{k-1}) \\
&= S(n) + 8 S(n) + 4 \sum_{k=0}^{2(n+1)} (-2)^{k-2} \binom{2n}{k-2} F_{k-2} - 4S(n) - 4 \sum_{k=0}^{2(n+1)} (-2)^{k-1} \binom{2n}{k-1} F_{k-1} \\
&= S(n) + 8 S(n) - 4 S(n) \\
&= 5 S(n). \\ 
\end{align*}$$
It is important to understand that while all of the sums appear to be taken over $\{0, \ldots, 2(n+1)\}$, in actuality some terms are zero because the corresponding binomial coefficient in the summand is zero.
Consequently, we have $S(n+1) = 5S(n)$ which with the initial condition $S(0) = (-2)^0 \binom{0}{0} F_1 = 1$, we get $S(n) = 5^n$ as claimed.

The initial condition part of the above proof leads to an interesting observation for the astute reader:  note that the identity only needs the fact that in the initial case, the Fibonacci number is $1$, and $F_n = F_{n-1} + F_{n-2}$.  Therefore, if we use the usual definition $F_0 = 0, F_1 = 1$, then the identity can be written with $$S(n) = \sum_{k=0}^{2n} (-2)^k \binom{2n}{k} F_{k+1}$$ or $$S(n) = \sum_{k=0}^{2n} (-2)^k \binom{2n}{k} F_{k+\color{red}{2}}.$$  In fact, it is a special case of the more general identity  $$S(n,m) = \sum_{k=0}^{2n} (-2)^k \binom{2n}{k} F_{k+m} = F_m 5^n.$$  Another nice corollary is that when $m = 0$, we immediately get $S(n,0) = 0$.
