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Sum based on Probability - ISI MMA 2018 Question 24

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 

Start with hints

Do you really need a hint? Try it first!

"Hint 1" : 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)."Hint 2" \(S_n = [ 1+ \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} +...........]\) \(  \frac{S_n}{3} = [ \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} +............] \) \( \\ \)   Now use this two to apply arithmatic-geometric series . "Hint 3" \([S_n - \frac{1}{3}S_n]\) = [ 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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