# Find the sum of $n$ terms of the series $(3\times 8)+(6\times 11)+(9\times 14).............$

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

Toolbox:
• $n^{th}$ term of an A.P$=a+(n-1)d$
• Sum of $n$ terms of any series, $S_n=\sum t_n$
• $\sum ( A+B)=\sum A+\sum B$
• $\sum kA=k.\sum A$ where $k$ is a constant.
• $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
• $\sum n=\large\frac{n(n+1)}{2}$
The given series is $(3\times 8)+(6\times 11)+(9\times 14).............$
Step 1
We have to find the $n^{th}$ term, $t_n$ of the series.
Each term consists of product of $2$ terms.
The first term of each forms a sequence $3,6,9,............$
This sequence is an A.P. with common difference $3$.
$\therefore$ The $n^{th}$ term of this sequence is $3+(n-1)3=3n$
$i.e.,$ The first term of $t_n$ is $3n$
Step 2
The second term of $t_n$ forms the sequence $8,11,14.........$
This sequence is also an A.P. with first term $a=8$ and $d=3$
The $n^{th}$ term of this sequence is $8+(n-1)3=3n+5$
$\therefore$ The $2^{nd}$ term of $t_n$ is $3n+5$
$\Rightarrow\:$ The $n^{th}$ term of the given series is
$t_n=3n.(3n+5)$
$\Rightarrow\:t_n=9n^2+15n$
Step 3
We know that Sum of $n$ terms of any series, $S_n=\sum t_n$
$\Rightarrow\:S_n=\sum (9n^2+15n)$
We know that $\sum ( A+B)=\sum A+\sum B$ and
$\sum kA=k.\sum A$
$\Rightarrow\:S_n=9\sum n^2+15\sum n$
We know that $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$ and
$\sum n=\large\frac{n(n+1)}{2}$
$\Rightarrow\:S_n=9.\large\frac{n(n+1)(2n+1)}{6}$$+15.\large\frac{n(n+1)}{2} \Rightarrow\:S_n=\large\frac{3n(n+1)}{2}$$\big[(2n+1)+5\big]$
$\Rightarrow\:S_n=\large\frac{3n(n+1)(2n+6)}{2}$
$\Rightarrow\:S_n=3n(n+1)(n+3)$