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# Sum of n terms of series $\;1.3.5+3.5.7+5.7.9+$.......is

$(a)\;n\;(2n^3+8n^2+7n-2)\qquad(b)\;n\;(8n^3+11n^2-n-3)\qquad(c)\;8n^3+12^2-2n-3\qquad(d)\;None\;of\;the\;above$

Answer : (a) $\;n\;(2n^3+8n^2+7n-2)$
Explanation :
$a_{r}=(2r-1)(2r+1)(2r+3)$
$=8r^3+12r^2-2r-3$
$S_{n}=\sum_{r=1}^{n}\;a_{r}$
$=8\;\sum_{r=1}^{n}\;r^3\;+12\;\sum_{r=1}^{n}\;r^2\;-\sum_{r=1}^{n}\;r-3n$
$=8\;[\frac{n(n+1)}{2}]^2+12\;\large\frac{n(n+1)(2n+1)}{6}-2\;[\frac{n(n+1)}{2}]-3n$
$=n\;(2n^3+8n^2+7n-2)\;.$
answered Jan 22, 2014 by 1 flag