Let $u(n)$ — the Fibonacci sequence. Prove that $$u(1)^3+...+u(n)^3=\frac{ 1 }{ 10 } \left[ u(3n+2)+(-1)^{n+1}6u(n-1)+5 \right].$$
I suppose we need prove that equality by iduction in $n$. It's obviously holds at $n=1$. So let $P(k)$ holds, then also must $P(k + 1)$ holds. What do we do next?
Consider the series \begin{align} \sum_{r=1}^{n} t^{r} = \frac{t \, (1-t^{n})}{1-t} \end{align} and \begin{align} F_{n}^{3} &= \frac{1}{5 \sqrt{5}} \, (\alpha^{n} - \beta^{n})^{3} \\ &= \frac{1}{5} \left( F_{3n} - 3 (-1)^{n} \, F_{n} \right), \end{align} since $\sqrt{5} \, F_{n} = \alpha^{n} - \beta^{n}$ and $2 \alpha = 1 + \sqrt{5}$, $2 \beta = 1 - \sqrt{5}$. Now, \begin{align} \sum_{r=1}^{\infty} F_{n}^{3} &= \frac{1}{5 \sqrt{5}} \, \left[ \frac{\alpha^{3} (1 - \alpha^{3n})}{1 - \alpha^{3}} - \frac{\beta^{3} (1 - \beta^{3n})}{1 - \beta^{3}} \right] - \frac{3}{5 \sqrt{5}} \, \left[ \frac{-\alpha (1 - (-1)^{n} \alpha^{n})}{1 + \alpha} - \frac{- \beta (1 - (-1)^{n} \beta^{n})}{1 + \beta} \right] \\ &= \frac{1}{10} \, \left( F_{3n+2} - 6 \, (-1)^{n} \, F_{n-1} + 6 - F_{2} \right) \\ &= \frac{1}{10} \, \left( F_{3n+2} - 6 \, (-1)^{n} \, F_{n-1} + 5 \right) \end{align}
