Fibonacci-Geometric Progression!

Let $x = \frac{1}{10} +\frac{1}{10^2}+ \frac{2}{10^3}+ \frac{3}{10^4}+\frac{5}{10^5}+\frac{8}{10^6}+ \cdots$

$\frac{x}{10} = \frac{1}{10^2} +\frac{1}{10^3}+ \frac{2}{10^4}+ \frac{3}{10^5}+\frac{5}{10^6}+\frac{8}{10^7}+ \cdots$

$x+\frac{x}{10} = \frac{1}{10}+ \frac{2}{10^2}+ \frac{3}{10^3}+\frac{5}{10^4}+\frac{8}{10^5}+ \cdots$

$x+ \frac{x}{10} + 1 = 1 + \frac{1}{10}+ \frac{2}{10^2}+ \frac{3}{10^3}+\frac{5}{10^4}+\frac{8}{10^5}+ \cdots$

$x+ \frac{x}{10} + 1 = 10 \cdot (\frac{1}{10} +\frac{1}{10^2}+ \frac{2}{10^3}+ \frac{3}{10^4}+\frac{5}{10^5}+\frac{8}{10^6}+ \cdots)$

$x+ \frac{x}{10} + 1 = 10x$

$x= \frac{10}{89}$

So, our answer is $10+89 = \boxed {99}$

Though, $\color{#20A900}{\text{Fibonacci Sequence}}$ is generally defined by recursive relation, but it can also be defined explicitly using $\color{#D61F06}{\large F_n\,=\,\frac{(1+\sqrt5)^{n}\,-\,(1-\sqrt5)^{n}}{2^{n} \cdot \sqrt5 \cdot }}$ , where $n\,\epsilon\,\mathbb{N} \,\,.............\,\, \color{#3D99F6}{\text{By Binet's Fibonacci Formula}}$ .

Now, the given series is equal to $\large \sum_{n=1}^{\infty} \frac{F_n}{10^{n}}\,=\,\sum_{n=1}^{\infty} \frac{(1+\sqrt5)^{n}\,-\,(1-\sqrt5)^{n}}{10^{n} \cdot 2^{n} \sqrt5}$ . $\implies \frac{1}{\sqrt5} \cdot \left(\sum_{n=1}^{\infty} \left(\frac{1+\sqrt5}{20}\right)^{n} \,-\,\sum_{n=1}^{\infty} \left(\frac{1-\sqrt5}{20}\right)^{n} \right)\,=\, \frac{1}{\sqrt5}\left(\frac{1+\sqrt5}{19-\sqrt5}\,-\,\frac{1-\sqrt5}{19+\sqrt5}\right)$ $=\frac{40\sqrt5}{356\sqrt5}\,=\,\frac{10}{89}$ . Hence $\color{#3D99F6}{a\,+\,b\,=\,10\,+\,89\,=\,99}$ .

Binet's Fibonacci Formula

This is a more or less a similar approach as @Manuel Kahayon used.

---

$\begin{aligned} S & = \frac{1}{10} +\frac{1}{10^2}+ \frac{2}{10^3}+ \frac{3}{10^4}+\frac{5}{10^5}+\frac{8}{10^6}+ \ldots \\ \frac{S}{10} & = \frac{1}{10^2} +\frac{1}{10^3}+ \frac{2}{10^4}+ \frac{3}{10^5}+\frac{5}{10^6}+\frac{8}{10^7}+ \ldots \\ S-\frac{S}{10} & = \frac{9S}{10} = \frac{1}{10}+\underbrace{ \frac{1}{10^3}+\frac{1}{10^4}+\frac{2}{10^5}+\frac{3}{10^6}+\frac{5}{10^7} +\ldots }_{\frac{S}{100}} \\ \frac{9S}{10} & =\frac{1}{10}+\frac{S}{100} \\ \therefore S & = \frac{10}{89} \\ & \Rightarrow 10+89 =\boxed{99} \end{aligned}$

Note that the generating function of the Fibonacci sequence is, $f(x)=\sum_{k=1}^{\infty}F_kx^k=\frac{x}{1-x-x^2}$

Hence, the given expression, is equal to $f(\frac{1}{10})=\frac{10}{89}$

$\text{Required answer}=\boxed{99}$

Consider a polynominal that

$x + x^2 + 2x^3 + 3x^4 + 5x^8 ⋯ = A$

Then

$(x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 ⋯ )(x + 1)$

$= x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + ⋯$

$=(A - x)/x$

Therefore

$A(x + 1) = (A - x)/x$

$Ax(x + 1) + x = A$

$A(x^2 + x -1) = -x$

$\therefore A = \frac{-x}{x^2 + x -1}$

In this problem, substitute 1/10 in x.

$\frac{-1/10}{1/100 + 10/100 - 100/100}$

$=\frac{10}{89}$

So the answer is $10 + 89 = 99$ .

Let the sum be $S$ , then we have $S-\frac{S}{10}=\frac{1}{10}+\frac{S}{100}$ and solve for $S$

$\sum_{n=1}^\infty F_n x^n = \frac{x}{1-x-x^2}$ generating function

put x = $\frac{1}{10}$

$\sum_{n=1}^\infty \frac{F_n}{10^n} =\frac{10}{89}$