An algebra problem by Saurabh Patil

Relevant wiki: Arithmetic-Geometric Progression

$\begin{aligned} S & = \frac {2+6}{4^{100}} + \frac {2+2(6)}{4^{99}} + \frac {2+3(6)}{4^{98}} + \cdots + \frac {2+99(6)}{4^2} + \frac {2+100(6)}4 \\ & = \frac 1{4^{100}} \left((2+6)+(2+2(6))4 + (2+3(6))4^2 + \cdots + (2+100(6))4^{99} \right) \\ & = \frac 1{4^{100}} \left(2 {\color{#3D99F6}\sum_{n=0}^{99} 4^n} + 6{\color{#D61F06}\sum_{n=1}^{100} n \cdot 4^{n-1}} \right) & \small {\color{#3D99F6}\text{GP}} \text{ and } \color{#D61F06} \text{AGP} \\ & = \frac 1{4^{100}} \left(2 \cdot {\color{#3D99F6}\frac {4^{100}-1}{4-1}} + 6 {\color{#D61F06}\left(\frac {1-100(4^{100})}{1-4} + \frac {4(1-4^{99})}{(1-4)^2}\right)} \right) \\ & = 200 \end{aligned}$

$\implies 3S = \boxed{600}$

Simple write general term and apply sumation you will get one AGP AND ONE GP solve them both and you get your answer