<s>[INST] Prove for all $n\in \mathbb{N}$ that $\sum_{i=0}^{n} i\cdot F_{2i} = (n+1)F_{2n + 1} - F_{2n + 2}$. $F_n$ denotes the Fibonacci sequence where $n$ is the term of the Fibonacci number in the sequence. ($F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, $F_4=3$, ... $F_n=F_{n-1} + F_{n-2}$)
I want to prove this using strong induction and have formulated base cases for when $n=0$ and when $n=1$.
My induction hypothesis is that for all $k\in \mathbb{N}$, $k\geq 1$ we can assume that for $k-1$, $\sum_{i=0}^{k-1} i\cdot F_{2i} = (k)F_{2k - 1} - F_{2k}$ and that for $k$, $\sum_{i=0}^{k} i\cdot F_{2i} = (k + 1)F_{2k + 1} - F_{2k + 2}$.
For my induction step I want to show that for $k+1$, $\sum_{i=0}^{k+1} i\cdot F_{2i} = (k+2)F_{2k + 3} - F_{2k + 4}$. I've been stuck on the equation below for the past couple of hours trying to figure out how I can derive the equation for $k + 1$. 
\begin{align*}
         \sum_{i=0}^{k+1} i\cdot F_{2i} &= (k + 1)\cdot F_{2k + 2} + \sum_{i=0}^{k} i\cdot F_{2i}\\
         &= (k + 1)\cdot F_{2k + 2} + (k + 1)F_{2k + 1} - F_{2k + 2} + \sum_{i=0}^{k - 1} i\cdot F_{2i}\\
         &= (k + 1)\cdot F_{2k + 2} + (k + 1)F_{2k + 1} - F_{2k + 2} + (k)F_{2k - 1} - F_{2k}\\
      \end{align*}
Any help is appreciated.
 [/INST] \begin{align}
\sum_{i=0}^{k+1}iF_{2i}&=(k+1)F_{2k+2}+(k+1)F_{2k+1}-F_{2k+2}\\
&=kF_{2k+2}+(k+1)(F_{2k+3}-F_{2k+2})\\
&=(k+1)F_{2k+3}-F_{2k+2}\\
&=(k+1)F_{2k+3}-(F_{2k+4}-F_{2k+3})\\
&=(k+2)F_{2k+3}-F_{2k+4}.
\end{align}
</s>