Q. The sum of the series $1 \bullet 2 \bullet 3 + 2 \bullet 3 \bullet 4 + 3 \bullet 4 \bullet 5 + ...$ to $n$ terms is
A) $n(n + 1)(n + 2)$
B) $(n + 1)(n + 2)(n + 3)$
C) $\frac{1}{4}n(n + 1)(n + 2)(n + 3)$
D) $\frac{1}{4}(n + 1)(n + 2)(n + 3)$
Explanation
(c)
Here, $T_{n} = n(n + 1)(n + 2)$
$= n^{3} + 3n^{2} + 2n$
$\therefore\ S_{n} = \sum_{}^{}T_{n}$
$= \left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{3} + \frac{3n(n + 1)(2n + 1)}{6} + \frac{2n(n + 1)}{2}$
$= \frac{1}{4}n(n + 1)(n + 2)(n + 3)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.