Sequences and Series 2 Question 11
12. The sum $V _1+V _2+\ldots+V _n$ is
(a) $\frac{1}{12} n(n+1)\left(3 n^{2}-n+1\right)$
(b) $\frac{1}{12} n(n+1)\left(3 n^{2}+n+2\right)$
(c) $\frac{1}{2} n\left(2 n^{2}-n+1\right)$
(d) $\frac{1}{3}\left(2 n^{3}-2 n+3\right)$
Show Answer
Answer:
Correct Answer: 12. $\frac{n^{2}(n+1)}{2}$
Solution:
- Here, $V _r=\frac{r}{2}[2 r+(r-1)(2 r-1)]=\frac{1}{2}\left(2 r^{3}-r^{2}+r\right)$
$\therefore \Sigma V _r=\frac{1}{2}\left[2 \Sigma r^{3}-\Sigma r^{2}+\Sigma r\right]$
$=\frac{1}{2} 2 \frac{n(n+1)}{2}^{2}-\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}$
$\Rightarrow \quad=\frac{n(n+1)}{12}[3 n(n+1)-(2 n+1)+3]$
$=\frac{1}{12} n(n+1)\left(3 n^{2}+n+2\right)$
