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Q)

# Find the sum of $n$ terms of the series $1\times 2+2\times 3+3\times 4+4\times 5+.............$

$\begin{array}{1 1} \large\frac{n(n+1)(n+2)}{3} \\ \large\frac{n(n+1)(n+2)}{6} \\ \large\frac{n(n+1)(2n+1)}{3} \\ \large\frac{n(n+1)(n+2)}{2} \end{array}$ Comment
A)
Toolbox:
• For any series $\sum t_n=Sum =S_n$
• $\sum n=1+2+3+.......n=\large\frac{n(n+1)}{2}$
• $\sum n^2=1^2+2^2+3^2+......n^2=\large\frac{n(n+1)(2n+1)}{6}$
• $\sum (A+B)=\sum A+\sum B$
The given series is $(1\times 2)+(2\times 3)+(3\times 4)+(4\times 5)+.........$
Step 1
We have to find the general term $t_n$ of the series.
The $n^{th}$ term of the series =$n^{th}$ bracket.
Each bracket consists of product of two terms.
The first term in each bracket are $(1,2,3............n)$
$\therefore$ The first term in $n^{th}$ bracket = $n$.
Similarly the $2^{nd}$ therm in each bracket are $(2,3,4......n+1)$
$\therefore$ the $2^{nd}$ term in $n^{th}$ bracket$=(n+1)$
$\Rightarrow\:n^{th}$ term of the series$=t_n=n\times (n+1)$
$i.e., \: t_n=n^2+n$
Step 2
We know that $\sum t_n=S_n$ (Sum of the series.)
$\Rightarrow\:$ Sum of the series $S_n=\sum t_n=\sum (n^2+n)$
Since we know that $\sum (A+B)=\sum A+\sum B$
$\Rightarrow\:S_n=\sum n^2+\sum n$
We know that $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$ and $\sum n=\large\frac{n(n+1)}{2}$
$\Rightarrow\:S_n=\large\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
Taking $\large\frac{n(n+1)}{2}$ common
$\Rightarrow\:S_n=\large\frac{n(n+1)}{2}\big(\frac{2n+1}{3}$$+1\big)$
$\Rightarrow\:S_n=\large\frac{n(n+1)}{2}.\frac{2n+1+3}{3}=\frac{n(n+1)(2n+4)}{6}$
$\therefore$ Sum of $n$ terms of the given series $=\large\frac{n(n+1)(n+2)}{3}$