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# Find the sum of the series 1.4 + 2.5 + 3.6----- upto 20 terms.

$\begin{array}{1 1} 2500 \\ 1560 \\ 8270 \\ 3500 \end{array}$

$Explanation : \;T_{n}=n(n+3)$
$=n^2+3n$
$S_{n}=\sum\;T_{n}=\sum\;n^2+3\;.\sum\;n$
$=\large\frac{n(n+1)(2n+1)}{6}+3\;.\large\frac{n(n+1)}{2}$
$S_{20}=\large\frac{20*21*41}{6}+3\;.\large\frac{20*21}{2}$
$=2870+630$
$=3500.$
edited Jan 24, 2014