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Sum based on Probability - ISI MMA 2018 Question 24

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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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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

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