Question

Find the sum to n terms of series $\mathrm{1.2.3}+\mathrm{2.3.4}+\mathrm{3.4.5}+\mathrm{4.5.6}+........$ .

• A. $\frac{n\times (n+1)\times (n+2)\times (n+3)}{4}$
• B. $\frac{(n+1)\times (n+2)\times (n+3)}{4}$
• C. $\frac{n\times (n+1)\times (n+2)\times (n+3)}{12}$
• D. $\frac{n\times (n+1)\times (n+2)\times (n+3)}{8}$

Answer 1

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

$a_n=n(n+1)(n+2)$

$S_n=\sum a_n$

$=\sum n(n+1)(n+2)$

$=\sum n(n^2+n+2n+2)$

$=\sum (n^3+3n^2+2n)$

$=\sum n^3+\sum 3n^2+\sum 2n$

$=\sum n^3+3\sum n^2+2\sum n$

$=\frac{n^2(n+1{)}^2}{4}+\frac{3n(n+1)(2n+1)}{6}+\frac{2n(n+1)}{2}$

$=\frac{n(n+1)}{2}[\frac{n(n+1)}{2}+2n+1+2]$

$=\frac{n(n+1)}{2}[\frac{n^2+n}{2}+2n+3]$

$=\frac{n(n+1)}{2}\left[\frac{n^2+n+4n+6}{2}\right]$

$=\frac{n(n+1)(n^2+5n+6)}{2\times 2}$

$=\frac{n(n+1)(n+2)(n+3)}{4}$