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# Find the sum of n terms of the series 1.5+3.8+5.11+---------------n

$(a)\;5n^2+6n+10\qquad(b)\;\frac{n(n+1)(2n+1)}{6}\qquad(c)\;\frac{4n^3+7n^2-n}{2}\qquad(d)\;6n^2+9n+1$

Answer : (c) $\frac{4n^3+7n^2-n}{2}$
Explanation : $T_{n}=(2n-1)(3n+2)$
$S_{n}=\sum\;T_{n}=\sum\;[6n^2+n-2]$
$=6\;\sum\;n^2+\sum\;n-\sum\;2$
$=\frac{6(n)(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-2n$
$=\frac{4n^3+6n^2+2n+n^2+n-4n}{2}$
$=\frac{4n^3+7n^2-n}{2}.$