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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Sequence and Series

Sum of n terms of series $(n^2-1^2)$ + $2\;.(n^2-2^2)$ + $3\;.(n^2-3^2)$+.....is

$(a)\;0\qquad(b)\;2n\;(n^2-1)\qquad(c)\;\frac{1}{4}\;n^2\;(n^2-1)\qquad(d)\;\frac{1}{4}\;n\;(n+1)^2$

1 Answer

Answer : (c) $\;\frac{1}{4}\;n^2\;(n^2-1)$
Explanation : $\;a_{r}=r\;(n-r)\;(n+r)$
$=n^2r-r^3$
$S_{n}=n^2\;\sum_{r=1}^{n}\;r-\sum_{r=1}^{n}\;r^3$
$=n^2\;\large\frac{n(n+1)}{2}-\large\frac{n^2(n+1)^2}{4}$
$=\large\frac{n^2(n+1)}{2}\;[n-\frac{n+1}{2}]$
$=\frac{1}{4}\;n^2\;(n^2-1)\;.$
answered Jan 22, 2014 by yamini.v
 
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