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
Topic
Probability
Difficulty Level
Medium
Suggested Book
A first course in Probability by Sheldon Ross
Start with hints
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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