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How can I Show a Fibonacci identity: $$\sum_{k=0}^{n}F_{k}F_{n-k}=\sum_{k=0}^{n}(k+1)F_{k+1}(-2)^{n-k}$$ Can anyone help me, i have no idea.

Alain Remillard
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Kathe
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  • There are always things you can do when you don't know how to tackle a problem; see https://math.meta.stackexchange.com/a/27933. – joriki May 02 '20 at 15:25
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    I got a different formula : $$ \sum_{k=0}^{n}{F_{k}F_{n-k}}=\frac{\left(n+1\right)F_{n-1}+\left(n-1\right)F_{n+1}}{5} $$ – CHAMSI May 02 '20 at 17:43

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You may use the ordinary generating function of Fibonacci numbers: $$ \sum_{n\geq 0}F_n x^n = \frac{x}{1-x-x^2} \tag{1}.$$ Letting $[x^m]\,g(x)$ be the coefficient of $x^m$ in the Maclaurin series of $g(x)$, we have $$ \sum_{k=0}^{n} F_k F_{n-k} = [x^n]\left(\frac{x}{1-x-x^2}\right)^2\tag{2} $$ and $$ \sum_{k=0}^{n}(k+1)F_{k+1}(-2)^{n-k} = [x^n]\frac{1}{1+2x}\left(\frac{x}{1-x-x^2}\right)' \tag{3}$$ so the LHS and RHS are different things.

Jack D'Aurizio
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