Fibonacci sequance

[asif e elahi]

Let {${F_{n}}$} be the fibonacci sequance.Find the sum $\frac{F_{0}}{6^{0}}+\frac{F_{1}}{6^{1}}+\frac{F_{2}}{6^{2}}...............$.

[*Mahi*]

\[F_n = \frac 1 {\sqrt 5} \{ (\frac {1+\sqrt 5} {2})^n - (\frac {1-\sqrt 5} {2})^n \}\]

[sowmitra]

Lemma:$\displaystyle \sum_{i=0}^{\infty}r^i=\frac{1}{1-r}$, when, $|r|<1$.

Let, $\mathcal S(k)=\displaystyle \sum^{\infty}_{i=0}\frac{F_i}{k^i}$,
where,$k\geq 2$, and, $F_i$ is the $i$-th Fibonacci Number.


$$\displaystyle \therefore \mathcal S(k)=\sum^{\infty}_{i=0}\frac{1}{k^i}\cdot \frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^i-(\frac{1-\sqrt{5}}{2})^i]$$

$$\displaystyle =\frac{1}{\sqrt{5}}\cdot [\sum^{\infty}_{i=0}(\frac{1+\sqrt{5}}{2k})^i-\sum^{\infty}_{i=0}(\frac{1-\sqrt{5}}{2k})^i]$$

$$\displaystyle =\frac{1}{\sqrt{5}}\cdot [\frac{1}{1-\frac{1+\sqrt{5}}{2k}}-\frac{1}{1-\frac{1-\sqrt{5}}{2k}}]$$

$$\displaystyle =\frac{2k}{\sqrt{5}}\cdot [\frac{1}{(2k-1)-\sqrt{5}}-\frac{1}{(2k-1)+\sqrt{5}}]$$

$$\displaystyle =\frac{2k}{\sqrt{5}}\cdot \frac{(2k-1)+\sqrt{5}-(2k-1)+\sqrt{5}}{(2k-1)^2-5} $$

$$\displaystyle =\frac{2k}{\sqrt{5}}\cdot \frac{2\sqrt{5}}{(2k-1)^2-5}$$

$$\displaystyle =\frac{4k}{(2k-1)^2-5}$$

So, $$\displaystyle \mathcal S(k)=\frac{4k}{(2k-1)^2-5}.$$
$$\displaystyle \therefore \mathcal S(6)=\frac{4\cdot 6}{(2\cdot 6-1)^2-5}=\boxed{\frac{6}{29}}.$$

Note:

The condition $$k\geq 2$$ is necessary, or else, the series $$\sum^{\infty}_{i=0}(\frac{1+\sqrt{5}}{2k})^i$$ won't converge.