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$$ \sum_{i=0}^{n} (F_{2i+1})^2 = \;?$$

I know that sum of squares of first $n$ Fibonacci numbers is $F_{n} \times F_{n+1}$.

Martin Sleziak
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Tahir Imanov
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  • According to what you (already) know: $(F_{2i+1})^2={F_{2i+1}}\cdot{F_{2i+2}}-{F_{2i}}\cdot{F_{2i+1}}$. – barak manos Jan 15 '15 at 13:44
  • Well, it won't work. Because I don't know $$\sum_{i=0}^{i=n} ( F_{2i+1} \times F_{2i+2} )$$ or $$\sum_{i=0}^{i=n} ( F_{2i} \times F_{2i+1} )$$ . – Tahir Imanov Jan 15 '15 at 14:06

2 Answers2

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Let $S_1=\displaystyle \sum_{i=0}^{n} F_{2i+1}^2$, $S_2=\displaystyle \sum_{i=1}^{n} F_{2i}^2$. First consider the formular: $$F_{2k-1}F_{2k+1}=F_{2k}^2+1$$ Therefore $$F_{2k-1}^2+F_{2k+1}^2=(F_{2k+1}-F_{2k-1})^2+2F_{2k+1}F_{2k-1}=3F_{2n}^2+2$$ Hence $$\begin{aligned} 2S_1&=F_1^2+F_{2n+1}^2+\displaystyle \sum_{i=1}^{n} (F_{2i-1}^2+F_{2i+1}^2)\\&=1+F_{2n+1}^2+3S_2+2n \end{aligned}$$ Also $$S_1+S_2=F_{2n+1}F_{2n+2}$$ Thus $$S_1=\frac{1}{5}\left(F_{2n+1}^2+3F_{2n+1}F_{2n+2}+2n+1\right)$$

Tien Kha Pham
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We know that: $$F_n = \frac{1}{\sqrt{5}}\left(\sigma^n - \bar{\sigma}^n\right)\tag{1}$$ where $\sigma,\bar{\sigma}$ are the roots of $x^2-x-1$. On the other hand, $$ L_n = \sigma^n+\bar{\sigma}^n, \tag{2}$$ and $\sigma\bar{\sigma}=-1$, so: $$ F_{2n+1}^2 = \frac{1}{5}\left(\sigma^{4n+2}+\bar{\sigma}^{4n+2}+2\right)=\frac{L_{4n+2}+2}{5}\tag{3}$$ and: $$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{5}\left(\sigma^2\frac{\sigma^{4(N+1)}-1}{\sigma^4-1}+\bar{\sigma}^2\frac{\bar{\sigma}^{4(N+1)}-1}{\bar{\sigma}^4-1}\right),$$ but since $\sigma^4-1 = (\sigma+1)^2-1 = \sigma(\sigma+2) $ the latter simplifies to: $$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{5}\left(\sigma\frac{\sigma^{4(N+1)}-1}{\sigma+2}+\bar{\sigma}\frac{\bar{\sigma}^{4(N+1)}-1}{\bar{\sigma}+2}\right),$$ or, since $(\sigma+2)(\bar{\sigma}+2)=5$, $$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{25}\left(\sigma(\bar{\sigma}+2)(\sigma^{4(N+1)}-1)+\bar{\sigma}(\sigma+2)(\bar{\sigma}^{4(N+1)}-1)\right),$$ $$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{25}\left(\sigma(\bar{\sigma}+2)(\sigma^{4(N+1)})+\bar{\sigma}(\sigma+2)(\bar{\sigma}^{4(N+1)})\right),$$ $$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{25}\left(2L_{4N+5}-L_{4N+4}\right),$$ $$\sum_{n=0}^{N}F_{2n+1}^2 = \color{red}{\frac{2}{5}(N+1)+\frac{1}{25}\left(L_{4N+5}+L_{4N+3}\right)}.\tag{4}$$

Jack D'Aurizio
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