Question

The sum to $n$ terms of the series $(1\times 2\times 3)+(2\times 3\times 4)+(3\times 4\times 5)+\dots$ is

• A. $\frac{n(n+1)(n+2)(2n+1)}{4}$
• B. $\frac{n(n+1)(n+2)(n+3)}{4}$
• C. $\frac{n(n-1)(n+1)(n+2)}{4}$
• D. $\frac{n(2n+1)(n+2)(n+3)}{4}$

Answer 1

The correct option is B $\frac{n(n+1)(n+2)(n+3)}{4}$

$T_n=n(n+1)(n+2)$
Let $S_n$ denote the sum to $n$ terms of the given series. Then,
$S_n=\sum _{k=1}^n{T}_k$
$=\sum _{k=1}^{n}k(k+1)(k+2)$
$=\sum _{k=1}^n(k^3+3k^2+2k)$
$=\left(\sum _{k=1}^n{k}^3\right)+3\left(\sum _{k=1}^n{k}^2\right)+2\left(\sum _{k=1}^{n}k\right)$
$=\left(\frac{n(n+1{)}^2}{2}\right)+\frac{3n(n+1)(2n+1)}{6}+\frac{2n(n+1)}{2}$
$=\frac{n(n+1)}{2}\{\frac{n(n+1)}{2}+(2n+1)+2\}$
$=\frac{n(n+1)}{4}\{n^2+n+4n+2+4\}$
$=\frac{n(n+1)}{4}(n^2+5n+6)$
$=\frac{n(n+1)(n+2)(n+3)}{4}$


Alternate Solution:
When $n=1$,
$S=1\times 2\times 3=6$
Now, checking the options,
$\frac{n(n+1)(n+2)(2n+1)}{4}=\frac{1\left(2\right)\left(3\right)\left(3\right)}{4}=\frac{9}{2}$
$\frac{n(n+1)(n+2)(n+3)}{4}=\frac{1\left(2\right)\left(3\right)\left(4\right)}{4}=6$
$\frac{n(n-1)(n+1)(n+2)}{4}=0$
$\frac{n(2n+1)(n+2)(n+3)}{4}=\frac{1\left(3\right)\left(3\right)\left(4\right)}{4}=9$

Hence, the correct option is
$\frac{n(n+1)(n+2)(n+3)}{4}$