Question

The value of the sum $\mathrm{1.2.3}+\mathrm{2.3.4}+\mathrm{3.4.5}+.......$ upto n terms=

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

Answer 1

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

Sum of $\mathrm{1.2.3}+\mathrm{2.3.4}+\mathrm{3.4.5}+.......$ upto $n$ terms

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

$S_n=$ Sum of $T_n$

$={\sum }_{n=1}^{n}n(n+1)(n+2)$

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

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

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

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

Hence, it is the correct answer.