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In this topic (Binomial coefficient and fibonacci numbers), it can be easily seen the sum of binomial coefficient and fibonacci numbers is

$$ \sum_{k=0}^{n} \binom{n}{k}F_k = F_{2n}. $$

I have also proved it via Binet formula of the Fibonacci numbers ( you can find the Binet formula in http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html).

Hereby, how can I find the following sum of the square of this summation formula

$$ \sum_{k=0}^{n} \left[ \binom{n}{k}F_k \right]^2 = ? $$

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Again using Binet's formula with $\phi:=\frac{1+\sqrt{5}}{2},\,\varphi:=\frac{1-\sqrt{5}}{2}$, the left-hand side is $$\frac{1}{5}\sum_{k=0}^n\binom{n}{k}^2(\phi^{2k}+\varphi^{2k}-2(-1)^k).$$In terms of Legendre functions of the first kind, according to Wolfram $$\sum_{k=0}^n\binom{n}{k}^2x^k=(1-x)^nP_n\left(\frac{1+x}{1-x}\right).$$The desired result simplifies to $$\frac{1}{5}\left((-\phi)^nP_n\left(-\sqrt{5}\right)+\phi^{-n}P_n\left(\sqrt{5}\right)-2^{n+1}P_n\left(0\right)\right).$$

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  • $\begingroup$ Thank you for a great answer. Can we represent your result with Fibonacci numbers? $\endgroup$ – drxy Apr 8 at 12:23

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