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 $$\frac{\left(-1\right)^{n}F_{2n+1}-\left(2n+1\right)}{5}$$
 A: Write $\varphi:=\tfrac{1+\sqrt{5}}{2}$ so $F_n=\tfrac{\varphi^n-(-\varphi^{-1})^n}{\sqrt{5}}$ in the convention $F_0=0,\,F_1=1$. Since $1+\varphi^2=\sqrt{5}\varphi$, the sum is$$\begin{align}\tfrac15\sum_{k=0}^n((-1)^k(\varphi^{2k}+\varphi^{-2k})-2)&=\frac{\tfrac{1-(-\varphi^2)^{n+1}}{1+\varphi^2}+\tfrac{1-(-\varphi^{-2})^{n+1}}{1+\varphi^{-2}}-2(n+1)}{5}\\&=\frac{-(-\varphi^2)^{n+1}-\varphi^2(-\varphi^{-2})^{n+1}-\sqrt{5}\varphi(2n+1)}{5\sqrt{5}\varphi}\\&=\frac{(-1)^n\varphi^{2n+1}+(-1)^n\varphi^{-2n-1}-\sqrt{5}(2n+1)}{5\sqrt{5}}\\&=\frac{(-1)^nF_{2n+1}-(2n+1)}{5}.\end{align}$$
A: $F_k=\frac{a^k-b^k}{\sqrt{5}}, a+b=1,ab=-1, a^2=a+1, b^2=b+1$
We will make multiple use of there relations below:
$$S=\frac{1}{5}\sum_{k=0}^{n} (-1)^k[a^{2k}+b^{2k}-2(ab)^k]~~~~(1)$$ $$S=\frac{1}{5}\sum_{k=0}^{n}(-1)^k\left[a^{2k}+b^{2k}-2(n+1)]\right]~~~~(2)$$
$$S=\frac{1}{5}\left[\frac{(-1)^{n+1}a^{2(n+1)}-1}{-a^2-1}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2-1}-2(n+1)\right]~~~~~(3)$$
$$S=\frac{1}{5}\left[\frac{(-1)^{n+1}a^{2(n+1)}-1}{-a^2+ab}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2+ab}-2(n+1)\right]~~~~~(4)$$
$$S=\frac{1}{5(a-b)}\left[\frac{(-1)^{n+1}a^{2(n+1)(a-b)}}{-a^2+ab}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2+ab}-2(n+1)(a-b)\right]~~~~(5)$$
$$=\frac{1}{5(a-b)}\left[(-1)^{n}[a^{2n+1}-b^{2n+1}]-1/a-1/b-2(n+1)(a-b)\right]$$
$$S=\frac{1}{5(a-b)}\left[(-1)^{n}[a^{2n+1}-b^{2n+1}]-(2n+1)(a-b)\right]~~~~(6)$$
$$\implies S=\frac{F_{2n+1}-(2n+1)}{5}~~~~~(7)$$
A: First let's take a look at the power series $ \sum\limits_{n\geq 0}{F_{n}^{2}x^{n}} $, it is easy to find what its radius of convergence would be using the ratio test, but for now let's denote it $ R<1 $, and let's define a function $ f:x\mapsto\sum\limits_{n=0}^{+\infty}{F_{n}^{2}x^{n}} $.
We have for all $ x\in\left(0,R\right) $ : \begin{aligned} \frac{1}{1-x}\sum_{n=0}^{+\infty}{F_{n}^{2}x^{n}}=\sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{F_{k}^{2}}\right)x^{n}}&=\sum_{n=0}^{+\infty}{F_{n}F_{n+1}x^{n}}\\&=\sum_{n=0}^{+\infty}{\left(F_{n+1}^{2}-F_{n}^{2}-\left(-1\right)^{n}\right)x^{n}}\\ \frac{f\left(x\right)}{1-x}&=\frac{f\left(x\right)}{x}-f\left(x\right)-\frac{1}{1+x}\\ \iff f\left(x\right)&=\frac{x\left(1-x\right)}{\left(1+x\right)\left(x^{2}-3x+1\right)}\end{aligned}
Now : \begin{aligned} \sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\left(-1\right)^{n-k}F_{k}^{2}}\right)x^{n}}&=\left(\sum_{n=0}^{+\infty}{\left(-1\right)^{n}x^{n}}\right)\left(\sum_{n=0}^{+\infty}{F_{n}^{2}x^{n}}\right)\\ &=\frac{x\left(1-x\right)}{\left(1+x\right)^{2}\left(x^{2}-3x+1\right)}\\ &=\frac{1-x}{5\left(x^{2}-3x+1\right)}+\frac{1}{5\left(x+1\right)}-\frac{2}{5\left(x+1\right)^{2}} \\ &=\frac{1}{5}\sum_{n=1}^{+\infty}{F_{n}^{2}x^{n-1}}+\frac{1}{5}\sum_{n=0}^{+\infty}{\left(-1\right)^{n}x^{n}}-\frac{2}{5}\sum_{n=1}^{+\infty}{n\left(-1\right)^{n}x^{n-1}}\\ &=\frac{1}{5}\sum_{n=0}^{+\infty}{\left(F_{n+1}^{2}+\left(-1\right)^{n}\left(2n+3\right)\right)x^{n}}\end{aligned}
Thus, for all $ n\in\mathbb{N} $ : $$ \sum_{k=0}^{n}{\left(-1\right)^{k}F_{k}^{2}}=\frac{\left(-1\right)^{n}F_{n+1}^{2}+2n+3}{5} $$
