Q39 / 66 · JEE Advanced 2007 (Paper 1) · Mathematics
Paragraph:
Let $V_r$ denotes the sum of the first $r$ terms of an arithmetic progression $(A P)$ whose first term is $r$ and the common difference is $(2 r-1)$. Let $T_r=V_{r+1}-V_r-2$ and $Q_r=T_{r+1}-T_r$ for $r=1,2, \ldots$Question:
The sum $V_1+V_2+\ldots+V_n$ is
- A$\frac{1}{12} n(n+1)\left(3 n^2-n+1\right)$
- Correct answerB$\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)$
- C$\frac{1}{2} n\left(2 n^2-n+1\right)$
- D$\frac{1}{3}\left(2 n^3-2 n+3\right)$
Solution
$$
\begin{aligned}
& V_r=\frac{r}{2}[2 r+(r-1)(2 r-1)]=\frac{1}{2}\left(2 r^3-r^2+r\right) \\
& \therefore \quad \Sigma V_r=\frac{1}{2}\left[2 \Sigma r^3-\Sigma r^2+\Sigma r\right] \\
& =\frac{1}{2}\left[2\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right] \\
& =\frac{1}{12} n(n+1)\left(3 n^2+n+2\right) \text {. } \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'B'.