Question

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

• A. $\frac{1}{6} n^{2}\left(2 n^{2}+1\right)$
• B. $\frac{1}{6}\left(n^{2}-1\right)(2 n-1)(2 n+3)$
• 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
$S=1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\ldots n$ term.
Now, $n$ th terms of the series,
$T_{n}=\{1+(n-1) \cdot 1\}\{2+(n-1) \cdot 1\}$
$\{3+(n-1) \cdot 1\}$
$\left[\because T_{n}=a+(n-1) d\right]$
$=(1+n-1)(2+n-1)(3+n-1)$
$=n(n+1)(n+2)$
$=n\left(n^{2}+2 n+n+2\right)$
$=n\left(n^{2}+3 n+2\right)$
$=n^{3}+3 n^{2}+2 n$
Now, $S=\Sigma T_{n}=\Sigma\left(n^{3}+3 n^{2}+2 n\right)$
$=\Sigma n^{3}+3 \Sigma n^{2}+2 \Sigma n$
$=\left[\frac{n(n+1)}{2}\right]^{2}+\frac{3 n(n+1)(2 n+1)}{6}+\frac{2 n(n+1)}{2}$
$=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+(2 n+1)+2\right]$
$=\frac{n(n+1)}{2}\left[\frac{n^{2}+n+4 n+2+4}{2}\right]$
$=\frac{n(n+1)\left(n^{2}+5 n+6\right)}{4}$
$=\frac{n(n+1)\left(n^{2}+2 n+3 n+6\right)}{4}$
$=\frac{n(n+1)[n(n+2)+3(n+2)]}{4}$
$=\frac{n(n+1)(n+2)(n+3)}{4}$