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# Find the sum of n terms of the series 2.4+5.7+8.10+------------

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

Can you answer this question?

Answer : (d) $\frac{n}{2}\;(6n^2+9n+1)$
Explanation : 2.4 +5.7 +8.10 +-------------
$T_{n}=(3n-1)(3n+1)$
$\sum\;T_{n}=S_{n}=\sum\;(9n^2-1)$
$S_{n}=9\;\sum\;n^2-\sum1$
$=\frac{9(n)(n+1)(2n+1)}{6}-n$
$S_{n}=\frac{n}{2} \; [3(n+1)(2n1)-2n ]$
$=\frac{n}{2}\;(6n^2+9n+1).$
answered Jan 3, 2014 by