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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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The sum of n terms of the infinite series $1.3^2$+$2.5^2$+$3.7^2$+-------$\infty$ is given by

$\begin{array}{1 1} \frac{n}{6(n+1)(6n^2+14n+7)} \\ \frac{n}{6(n+1)(2n+1)(3n+1)} \\4n^3+4n^2+n \\ [\frac{n(n+1)}{2}]^2\end{array}$

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Answer : (a) $\;\frac{n}{6(n+1)(6n^2+14n+7)}$
The $\;n^{th}\;$ term of the series is given by $\;n.(2n+1)^2$
Sum of n terms of the series is $\;\sum[\;n.(2n+1)^2]$
Expanding the term in the series we get
Sum S = $\;\sum\;(4n^3+4n^2+n)$
Separating the terms
$\sum\;4n^3=\;4*\;[n(n+1)^2/2]$
$\sum\;4n^2=\;4*\;n/6(n+1)(2n+1)$
$\sum\;n=\;n(n+1)/2$
$S=4*\;[n(n+1)^2/2]+4*n/6(n+1)(2n+1)\;+n(n+1)/2$
Taking n (n+1) common and expanding the terms
$S=n(n+1)\;[n(n+1)+2/3(2n+1)+1/2]$
$S=n(n+1)\;[n^2+n+4/3n+2/3+1/2]$
$S=n(n+1)\;[n^2+7/3n+7/6]$
Taking $\;1/6\;out\;of\;the\;2^{nd}\;part$
$S=\frac{n}{6(n+1)(6n^2+14n+7)}.$
answered Jan 17, 2014 by yamini.v
 

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