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Find the sum of $n$ terms of the series $3\times 1^2+5\times 2^2+7\times 3^2+.........$

$\begin{array}{1 1}\large\frac{n(n+1)(3n^2+5n+1)}{3} \\ \large\frac{n(n+1)(3n^2+5n+1)}{6} \\\large\frac{n(n+1)(3n^2+5n-1)}{6} \\ \large\frac{n(n+1)(3n^2+5n-1)}{3} \end{array} $

1 Answer

  • $n^{th}$ term of an A.P.$=a+(n-1)d$
  • Sum of $n$ terms of any series with $n^{th}$ term $t_n=S_n=\sum t_n$
  • $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
  • $\sum n^3=\large\frac{n^2(n+1)^2}{4}$
  • $\sum (A+B)=\sum A+\sum B$
  • $\sum k.A=k.\sum A$ where $k$ is a constant.
The Given series is $(3\times 1^2)+(5\times 2^2)+(7\times 3^2)+.........$
Step 1
We have to find the $n^{th}$ term $t_n$ of the series.
$n^{th}$ term of the series = $n^{th}$ bracket.
Each bracket consists of product of two terms.
The first term in each bracket forms a sequence $3,5,7.........$
This sequence is an A.P. with first term $a=3$ and common difference $d=2$
$\therefore$ $n^{th}$ term of this sequence $=a+(n-1)d$
$i.e., $The first term of the $n^{th}$ bracket $=2n+1$
Step 2
The $2^{nd}$ term of each bracket forms the sequence $1^2,2^2,3^2......n^2$
$i.e.,$ The $2^{nd}$ term of $n^{th}$ bracket $=n^2$
$\Rightarrow\:n^{th}$ term of the given series $t_n=(2n+1).n^2=2n^3+n^2$
Step 3
Sum of $n$ terms of any series with $n^{th}$ term $t_n=S_n=\sum t_n$
$\Rightarrow\:$ Sum of $n$ terms =$S_n=\sum t_n=\sum (2n^3+n^2)$
We know that $\sum (A+B)=\sum A+\sum B$ and
$\sum k.A=k.\sum A$ where $k$ is a constant.
$\Rightarrow\: S_n=2.\sum n^3+\sum n^2$
Also we know that $\sum n^3=\large\frac{n^2(n+1)^2}{4}$ and $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
Taking $\large\frac{n(n+1)}{2}$ common
$\Rightarrow\:$ Sum of $n$ terms of the given series $S_n=\large\frac{n(n+1)(3n^2+5n+1)}{6}$
answered Mar 7, 2014 by rvidyagovindarajan_1

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