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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

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

1 Answer

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
Taking n (n+1) common and expanding the terms
Taking $\;1/6\;out\;of\;the\;2^{nd}\;part$
answered Jan 17, 2014 by yamini.v
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