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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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Sum of n terms of series $\;S=1^2+2(2)^2+3^2+2(4)^2+5^2+\;....\;$ when n is even is :

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

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Answer : (d) $\;\frac{1}{2}n(n+1)^2$
Explanation : $Let \;n=2m$
$S=\;[1^2+2^2+3^2\;...\;+(2m)^2]+\;[2^2+4^2+6^2+\;...\;(2m)^2]$
$=\frac{1}{6}\;(2m)(2m+1)(4m+1)+\frac{4}{6}\;(m)(m+1)(2m+1)$
$=\frac{1}{3}m(2m+1)\;[4m+1+2m+2]$
$=\frac{1}{2}n(n+1)^2\;.$
answered Jan 22, 2014 by yamini.v
 
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