Question

The value of the sum $1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+\dots$ upto $n$ terms is equal to

• A. $\frac{1}{6}{n}^2\left(2n^2+1\right)$
• B. $\frac{1}{6}\left(n^2-1\right)(2n-1)(2n+3)$
• C. $\frac{1}{8}\left(n^2+1\right)\left(n^2+5\right)$
• D. $\frac{1}{4}n(n+1)(n+2)(n+3)$

Answer 1

Answer: (D)

Let given series be
$S=1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+\dots n$ term.
Now, $n$ th terms of the series,
$T_n=\{1+(n-1)\cdot 1\}\{2+(n-1)\cdot 1\}$
$\{3+(n-1)\cdot 1\}$
$\left[\because T_n=a+(n-1)d\right]$
$=(1+n-1)(2+n-1)(3+n-1)$
$=n(n+1)(n+2)$
$=n\left(n^2+2n+n+2\right)$
$=n\left(n^2+3n+2\right)$
$=n^3+3n^2+2n$
Now, $S=\Sigma T_n=\Sigma \left(n^3+3n^2+2n\right)$
$=\Sigma n^3+3\Sigma n^2+2\Sigma n$
$={\left[\frac{n(n+1)}{2}\right]}^2+\frac{3n(n+1)(2n+1)}{6}+\frac{2n(n+1)}{2}$
$=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+(2n+1)+2\right]$
$=\frac{n(n+1)}{2}\left[\frac{n^2+n+4n+2+4}{2}\right]$
$=\frac{n(n+1)\left(n^2+5n+6\right)}{4}$
$=\frac{n(n+1)\left(n^2+2n+3n+6\right)}{4}$
$=\frac{n(n+1)[n(n+2)+3(n+2)]}{4}$
$=\frac{n(n+1)(n+2)(n+3)}{4}$