sum of a Fibonacci type series Sum of $n$ (or in particular, $52$) terms of $1,6,7,13,20,33\ldots$
just cant think of any relation. do we have a direct formula for the sum of fibonacci numbers?
 A: Hint:  You can solve the recurrence by the standard techniques.  Assume a solution of the form $a_n=br^n$ and substitute into the recurrence $a_n=a_{n-1}+a_{n-2}$  The roots are the same as the Fibonacci series, but the constants are different.  The sum of the first $52$ terms becomes the sum of two geometric series.  Do you know how to sum those?
A: Your sequence satisfies the Fibonacci recurrence, but with $x_0=1$ and $x_1=6$. Thus, it has the closed form $x_n=A\varphi^n+B\widehat\varphi^n$, where $\varphi=\frac12(1+\sqrt5)$, $\widehat\varphi=\frac12(1-\sqrt5)$, $1=x_0=A+B$, and $6=x_1=A\varphi+B\widehat\varphi$. Solving the last two equations for $A$ and $B$, we find that
$$B=\frac{\varphi-6}{\varphi-\widehat\varphi}=\frac{\varphi-6}{\sqrt5}=\frac1{10}\left(5-11\sqrt5\right)$$
and
$$A=1-B=\frac1{10}(5+11\sqrt5)\;.$$
You can use now write
$$s_n=\sum_{k=0}^{n-1}x_k=A\sum_{k=0}^{n-1}\varphi^k+B\sum_{k=0}^{n-1}\widehat\varphi^k$$ 
and sum the geometric series to get a closed form for $s_n$ in terms of $\varphi$ and $\widehat\varphi$. If you wish, you can then manipulate this closed form a bit to discover that $x_n=6x_n-29F_n-6$ for each $n$. (This can also be proved by induction on $n$ if you actually suspect such a relationship.) I’ve left the detailed calculations spoiler-protected below.

 $$\begin{align*}\sum_{k=0}^{n-1}x_k&=A\sum_{k=0}^{n-1}\varphi^k+B\sum_{k=0}^{n-1}\widehat\varphi^k\\&=A\frac{1-\varphi^n}{1-\varphi}+B\frac{1-\widehat\varphi^n}{1-\widehat\varphi}\\&=\frac{A}{\widehat\varphi}\left(1-\varphi^n\right)+\frac{B}{\varphi}\left(1-\widehat\varphi^n\right)\\&=-\frac15\left(15+4\sqrt5\right)\left(1-\varphi^n\right)-\frac15\left(15-4\sqrt5\right)\left(1-\widehat\varphi^n\right)\\&=-6+\frac15\left(15+4\sqrt5\right)\varphi^n+\frac15\left(15-4\sqrt5\right)\widehat\varphi^n\\&=-6+\left(6A-\frac{29}{\sqrt5}\right)\varphi^n+\left(6B+\frac{29}{\sqrt5}\right)\widehat\varphi^n\\&=-6+6x_n-29\left(\frac{\varphi^n-\widehat\varphi^n}{\sqrt5}\right)\\&=-6+6x_n-29F_n\end{align*}$$

