Mysterious Sum

Let the value of the infinite sum be $S$ , which can be represented by $S = \dfrac{a}{b}$ , then we have:

$\begin{aligned} S & = 1+0.2 + 0.03 + 0.004 +... \\ & = \sum_{n=1}^\infty nx^{n-1} \quad \quad \small \color{#3D99F6}{\text{where }x = 0.1} \\ & = \sum_{n=1}^\infty \frac{dx^n}{dx} = \frac{d}{dx} \sum_{n=1}^\infty x^n = \frac{d}{dx} \left(\frac{x}{1-x}\right) \\ & = \frac{1}{1-x} + \frac{x}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2} \\ & = \frac{1}{(1-0.1)^2} = \frac{1}{0.81} = \frac{100}{81} \end{aligned}$

$\Rightarrow a + b = 100 + 81 = \boxed{181}$

Sarja onAP - GP . Käyttää kaavan avulla voidaan ratkaista

DUHHHH

Using binomial theoram given expression is (1-0.1)^2= 100/81