# Find the sum of n terms of the series 2.4 + 5.6 + 8.8 + 11.10 +---------

$(a)\;2n^2+5n+1\qquad(b)\;(6n^2+n+1)n\qquad(c)\;n(2n^2+5n+1)\qquad(d)\;n^2+2n+1$

Answer : (c) $n(2n^2+5n+1)$
Explanation : $T_{n}=(3n-1)(2n+2)$
$=6n^2+6n-2n-2=6n^2+4n-2$
$S_{n}=\sum\;T_{n}=6\;\sum\;n^2+4\;\sum\;n-\sum\;2$
$S_{n}=\frac{6n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-2n$
$=n(n+1)(2n+1)+2n(n+1-1)$
$=n(2n^2+5n+1).$