# Understand the problem

The sum of Infinity series $$1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + …….$$ is
##### Source of the problem

Sample Questions(MMA) : 2018 Problem no 24

Probability
Medium
##### Suggested Book
A first course in Probability by Sheldon Ross

Do you really need a hint? Try it first!

Do you know about Arithmatics-Geometric series .(A geometric series is the sum of a geometric sequence. Thus, with the series you just see if the relationship between the terms is arithmetic (each term increases or decreases by adding a constant to the previous term ) or geometric (each term is found by multiplying the previous term by a constant).
$$S_n = [ 1+ \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} +………..]$$ $$\frac{S_n}{3} = [ \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} +…………]$$ $$\\$$   Now use this two to apply arithmatic-geometric series .
$$[S_n – \frac{1}{3}S_n]$$ = [ 1 +$$\frac{1}{3}$$ + $$\frac{4}{3^2}$$ + $$\frac{4}{3^3}$$ + $$\frac{4}{3^4}$$+……….] $$\Rightarrow \frac{2}{3}S_n$$ = $$\frac{4}{3}[ 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + …………..]$$ = $$\frac{4}{3}$$ *$$\frac{1}{1-\frac{1}{3}}$$ $$\Rightarrow \frac{2}{3}S_n$$ = $$\frac{4}{3} * \frac{3}{2}$$ =2 $$\Rightarrow S_n = 3$$

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