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The sum of n terms of the series

\frac{5}{1.2}*\frac{1}{3}+\frac{7}{2.3}*\frac{1}{3^{2}}+\frac{9}{3.4}*\frac{1}{3^{3}}+\frac{11}{4.5}*\frac{1}{3^{4}}.......is

Option: 1

1-\frac{1}{2^{n-1}}*\frac{1}{3^{n}}


Option: 2

1+\frac{1}{n+1}*\frac{1}{3^{n}}


Option: 3

1-\frac{1}{n+1}*\frac{1}{3^{n}}


Option: 4

1+\frac{1}{2n-1}*\frac{1}{3^{n}}


Answers (1)

best_answer

As we learn

Telescopic Series

Let a_{1},a_{2},a_{3},------a_{n} be a sequence,

the expression  \sum_{r=1}^{n-1}\left ( a_{r}-a_{r+1} \right )= a_{1}-a_{n} is called Telescopic Series.
 

- wherein

e.g.

= \sum_{r=1}^{n}r\cdot r!


  = \sum_{r=1}^{n} ( r+1-1 ) r!

= \sum_{r=1}^{n}\left ( r+1 \right )!- r!

= \left ( n+1 \right )!- 1

 

In this Question,

 t_{n}=\frac{2n+3}{n(n+1)}*\frac{1}{3^{n}}=(\frac{3}{n}-\frac{1}{n+1})*\frac{1}{3^{n}}=\frac{1}{n*3^{n-1}}-\frac{1}{(n+1)3^{n}}

Put n=1, 2,3...n

\\S_{n}=\left (\frac{1}{1*3^{1-1}}-\frac{1}{(1+1)3^{1}} \right )+\left (\frac{1}{2*3^{2-1}}-\frac{1}{(2+1)3^{2}} \right )+\left (\frac{1}{3*3^{3-1}}-\frac{1}{(3+1)3^{3}} \right )\ldots\dots\left (\frac{1}{n*3^{n-1}}-\frac{1}{(n+1)3^{n}} \right )

\sum t_{n}=1-\frac{1}{n+1}*\frac{1}{3^{n}}

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Riya

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