How to find closed form of summation of Fibonacci Sequence? I created two formulas to prove a binary theory involving the Fibonacci sequence.
(1) $\sum_{i=0}^n F_{2i+1} $ 
Equation (1) is the sum of all Fibonacci numbers up to $F_n$ where every $i$ in $F_i$ is an odd number. 
I had come up with a closed form where (1) = $F_{n+1} -1$
Is this right? If so, how do I prove it?

The Fibonacci numbers have the form $(\alpha - \beta) F_{n} = \alpha^{n} - \beta^{n}$, where $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$. Now
\begin{align}
\sum_{k=0}^{n} (-1)^{k} F_{k} &= \frac{1}{\alpha - \beta} \sum_{k=0}^{n} \left( (- \alpha)^{k} - (- \beta)^{k} \right) \\
&= \frac{1}{\alpha - \beta} \left( \frac{1 - (-\alpha)^{n+1}}{1+\alpha} - \frac{1 - (-\beta)^{n+1}}{1+ \beta} \right) \\
&= \frac{1}{\alpha - \beta} \left( (-1)^{n} (\alpha^{n-2} - \beta^{n-2}) - (\alpha^{2} - \beta^{2}) \right) \\
&= (-1)^{n} F_{n-2} - (\alpha + \beta) \\
&= (-1)^{n} F_{n-2} -1.
\end{align} 
Since $F_{0} = 0$ the summation can also be seen as
\begin{align}
\sum_{k=1}^{n} (-1)^{k} F_{k} = (-1)^{n} F_{n-2} - 1.
\end{align}
