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# Sum based on Probability – ISI MMA 2018 Question 24

This is an interesting and cute sum based on the concept of Arithematic and Geometric series .The problem is to find a solution of a probability sum.

# 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 .(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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#### College Mathematics Program

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