$F_{2n} = F_{2n-2}+2F_{2n-4}+\dots+n$ rigorous proof Let $F_{n}$ be n-th fibonacci number($F_{0}$ = 0) and $g_{n} = F_{2n}$ if $n > 0$ $g_{0} = 1$.
I want to prove that $g_{n} = g_{n-1}+2g_{n-2}+\dots +ng_{0}$. It's obviously seen from direct evaluation but it's not so rigorous and my attempts to use induction end in failure..

There are many ways to prove this. One way is to notice that
$$
F_{2n} = F_{2n-1} + F_{2n-2} = 2F_{2n-2} + F_{2n-3} = 2F_{2n-2} + F_{2n-3} + F_{2n-4} - F_{2n-4} = 3F_{2n-2} - F_{2n-4}.
$$
Therefore
$$ g_n = 3g_{n-1} - g_{n-2}. $$
Given this, it is easy to prove the claim by induction. Let's rephrase it slightly:
$$
g_n = \sum_{i=1}^{n-1} ig_{n-i} + n.
$$
In particular, when $n = 0$ we get $g_0 = 0$, and when $n = 1$ we get $g_1 = 1$, both obviously correct. Assuming the claim holds for $n-1,n-2$, we have
$$
\begin{align*}
g_n &= 3g_{n-1} - g_{n-2} \\ &=
3\sum_{i=1}^{n-2} ig_{n-1-i} + 3(n-1) - \sum_{i=1}^{n-3} ig_{n-2-i} - (n-2) \\ &=
\sum_{i=1}^{n-2} i(3g_{n-1-i} - g_{n-2-i}) + 2n-1 \\ &=
\sum_{i=1}^{n-2} ig_{n-i} + 2n-1 \\ &=
\sum_{i=1}^{n-1} ig_{n-i} - (n-1)g_1 + 2n-1 \\ &=
\sum_{i=1}^{n-1} ig_{n-i} + n.
\end{align*}
$$
