Sum of $i$ times $(i-1)^\text{th}$ Fibonacci Number

Question

Consider the expression $$\sum\limits_{i=1}^n i \cdot F_{i-1}$$, where $F_{0}=0, F_{1}=1, F_{2}=1, F{3}=2,$ etc. Is there a closed formula for this? If so, find it.

Answer 1

Take the generating function of $F_n$ (which is absolutely convergent for $|x|<\frac{2}{\sqrt{5}+1}$: $$ \frac{x}{1-x-x^2}=\sum_{i=0}^{\infty}F_i x^i\Rightarrow \frac{1}{1-x-x^2}=\sum_{i=1}^{\infty}F_i x^{i-1} $$ Take derivatives of both sides: $$ \frac{x^2+1}{\left(1-x-x^2\right)^2}=\sum_{i=1}^{\infty}i F_i x^{i-1}\Rightarrow $$ $$ \sum_{i=1}^{\infty}i F_i x^{i-1}+\sum_{i=1}^{\infty}F_i x^{i-1}=\sum_{i=1}^{\infty}(i+1)F_i x^{i-1}=\frac{2-x}{(1-x-x^2)^2} $$ And take the Taylor-series at $x=0$.