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?
 A: This proof uses only the definition.
\begin{align*}
F_{2n + 2} &= F_{2n + 1} + F_{2n} \\
&= F_{2n + 1} + F_{2n - 1} + F_{2(n-1)} \\
&= F_{2n + 1} + F_{2(n-1) + 1} + F_{2(n-2) + 1} + F_{2(n-2)} \\
&= \cdots \\
&= F_{2(n-k)} + \sum_{j = 0}^{k} F_{2(n-j) + 1} 
\end{align*}
Setting $k = n$ in the last line, we obtain the desired formula $F_{2n + 2} = 1 + \sum_{j = 0}^n F_{2j + 1}$.
A: 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}
A: We have that
$$
F_n=\frac{1}{2}\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}
+\frac{1}{2}\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}=\frac{1}{2}\big(\mu^{n-1}+\nu^{n-1}\big),
$$
and hence
$$
s=\sum_{i=0}^n F_{2i+1}=\frac{1}{2}\sum_{j=0}^n\left(\mu^{2j}+\nu^{2j}\right)=\frac{1}{2}
\left(\frac{\mu^{2n+2}-1}{\mu^2-1}+\frac{\nu^{2n+2}-1}{\nu^2-1}\right).
$$
Next
$$
\mu^2-1=\frac{6+2\sqrt{5}}{4}-1=\frac{1+\sqrt{5}}{2}=\mu,\quad \nu^2-1=\cdots=\nu,
$$
while
$$
\frac{1}{\mu}=\frac{2}{1+\sqrt{5}}=\frac{2\sqrt{5}-2}{4}=\nu.
$$
Hence
$$
s=\frac{1}{2}\left(\mu^{2n+1}+\nu^{2n+1}-\frac{1}{\mu}-\frac{1}{\nu}\right)=F_{2n+2}-1.
$$
A: Let's start with 
$$(1)\qquad \sum_{i=0}^NF_{2i+1}$$
As said in other answer, the Fibonnaci numbers have the form 
$$F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$$
where $\alpha$ and $\beta$ are the roots of the polinomial $x^2-x-1=0$, so $\alpha^2-1=\alpha$ and $\beta^1-1=\beta$.
Then
$$\begin{array}{rcl}
\sum_{i=0}^NF_{2i+1}&=&\frac{1}{\alpha-\beta}\sum_{i=0}^N\big(\alpha^{2i+1}-\beta^{2i+1}\big)\\
&=&\frac{1}{\alpha-\beta}\Bigg(\alpha\sum_{i=0}^N\big(\alpha^2\big)^i-\beta\sum_{i=0}^N\big(\beta^2\big)^i\Bigg)\\
&=&\frac{1}{\alpha-\beta}\Bigg(\alpha\frac{\big(\alpha^2\big)^{N+1}-1}{\alpha^2-1}-\beta\frac{\big(\beta^2\big)^{N+1}-1}{\beta^2-1}\Bigg)\\
&=&\frac{1}{\alpha-\beta}\Big(\big(\alpha^2\big)^{N+1}-1-\big(\beta^2\big)^{N+1}+1\Big)\\
&=&\frac{1}{\alpha-\beta}\Big(\alpha^{2N+2}-\beta^{2N+2}\Big)\\
&=&F_{2N+2}
\end{array}$$
For the second one you originally posted, "Sum the Fibonacci numbers $F_i$ with $i$ even": 
$$\begin{array}{rcl}
\sum_{i=1}^NF_{2i}&=&\frac{1}{\alpha-\beta}\sum_{i=1}^N\big(\alpha^{2i}-\beta^{2i}\big)\\
&=&\frac{1}{\alpha-\beta}\Bigg(\sum_{i=0}^N\big(\alpha^2\big)^i-\sum_{i=0}^N\big(\beta^2\big)^i\Bigg)\\
&=&\frac{1}{\alpha-\beta}\Bigg(\frac{\big(\alpha^2\big)^{N+1}-\alpha^2}{\alpha^2-1}-\frac{\big(\beta^2\big)^{N+1}-\beta^2}{\beta^2-1}\Bigg)\\
&=&\frac{1}{\alpha-\beta}\Bigg(\frac{\alpha^{2N+2}-\alpha^2}{\alpha}-\frac{\beta^{2N+2}-\beta^2}{\beta}\Bigg)\\
&=&\frac{1}{\alpha-\beta}\Big(\alpha^{2N+1}-\alpha-\beta^{2N+1}+\beta\Big)\\
&=&\frac{1}{\alpha-\beta}\Big(\alpha^{2N+1}-\beta^{2N+1}\Big)-1\\
&=&F_{2N+1}-1
\end{array}$$
