Closed form for $\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$ How to get a closed form for the following sum:
$$\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$$
Where $F_k$ is the $k$th Fibonacci number.
I tried to get a recurrence relation,but I failed and the closed form is given by $$\frac{\left(-1\right)^{n}F_{2n+1}-\left(2n+1\right)}{5}$$

$F_k=\frac{a^k-b^k}{\sqrt{5}}, a+b=1,ab=-1, a^2=a+1, b^2=b+1$
We will make multiple use of there relations below:
$$S=\frac{1}{5}\sum_{k=0}^{n} (-1)^k[a^{2k}+b^{2k}-2(ab)^k]~~~~(1)$$ $$S=\frac{1}{5}\sum_{k=0}^{n}(-1)^k\left[a^{2k}+b^{2k}-2(n+1)]\right]~~~~(2)$$
$$S=\frac{1}{5}\left[\frac{(-1)^{n+1}a^{2(n+1)}-1}{-a^2-1}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2-1}-2(n+1)\right]~~~~~(3)$$
$$S=\frac{1}{5}\left[\frac{(-1)^{n+1}a^{2(n+1)}-1}{-a^2+ab}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2+ab}-2(n+1)\right]~~~~~(4)$$
$$S=\frac{1}{5(a-b)}\left[\frac{(-1)^{n+1}a^{2(n+1)(a-b)}}{-a^2+ab}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2+ab}-2(n+1)(a-b)\right]~~~~(5)$$
$$=\frac{1}{5(a-b)}\left[(-1)^{n}[a^{2n+1}-b^{2n+1}]-1/a-1/b-2(n+1)(a-b)\right]$$
$$S=\frac{1}{5(a-b)}\left[(-1)^{n}[a^{2n+1}-b^{2n+1}]-(2n+1)(a-b)\right]~~~~(6)$$
$$\implies S=\frac{F_{2n+1}-(2n+1)}{5}~~~~~(7)$$
