Question

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

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

Answer 1

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

D. $\frac{1}{4}n(n+1)(n+2)(n+3)$
$\mathrm{1.2.3}+\mathrm{2.3.4}+....$ upto n term
$t_n=n(n+1)(n+2)$
$S_n=\Sigma t_n=\frac{1}{4}n(n+1)(n+2)(n+3)$