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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If $S_{n}=\frac{1}{6}\;n\;(2n^2+9n+13)\;$, then $\sum_{r=1}^n\;\sqrt{a_{r}}\;$ equals.

$(a)\;\frac{1}{2}n(n+3)\qquad(b)\;\frac{1}{2}n(n+5)\qquad(c)\;\frac{1}{2}n(n+2)\qquad(d)\;\frac{1}{2}n(n+1)$

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Answer : (a) $\frac{1}{2}n(n+3)$
Explanation : $a_{n}=S_{n}-S_{n-1}$
$=\frac{1}{6}n(2n^2+9n+13)-\frac{1}{6}(n-1)\;[2(n-1)^2+9(n-1)+13]$
$=\frac{1}{6}\;(6n^2+12n+6)$
$=(n+1)^2$
$\sqrt{a_{n}}=(n+1)$
$\sum_{r=1}^n\;\sqrt{a_{r}}=\sum_{r=1}^n\;r+1$
$=\frac{n(n+1)}{2}+n$
$=\frac{1}{2}\;n(n+3).$
answered Jan 20, 2014 by yamini.v
 

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