Question

Let $V_r$ denote the sum of the first $r$ terms of an arithmetic progression (AP), whose first term is $r$ and the common difference is $(2r-1)$. Let $T_r=V_r+1–V_r-2$ and $Q_r=T_r+1–T_{r}forr=1,2\dots .$The sum$V{1}_1+V_2+\dots +V_n$ is

• A. $\left(\frac{1}{12}\right)n(n+1)(3n^2+n+2)$
• B. $\left(\frac{1}{2}n\right)(2n^2-n+1)$
• C. $\left(\frac{1}{3}\right)(2n^3-2n+3)$
• D. None of these

Answer 1

The correct option is A

$\left(\frac{1}{12}\right)n(n+1)(3n^2+n+2)$

Explanation for the correct option:

Given $a=r$

$d=2r-1$

$S_n=\left(\frac{n}{2}\right)(2a+(n-1\left)d\right)$

$$\begin{gathered} V_r=\left(\frac{r}{2}\right)(r+(r-1\left)\right(2r-1) \\ =(\frac{1}{2}\left)\right(2r^3-r^2+r) \end{gathered}$$

$$\begin{gathered} \sum V_r=V_1+V_2+V_3+\dots \\ \sum V_r=\frac{1}{2}(2\sum r^3-\sum r^2+\sum r) \\ =\left(\frac{1}{2}\right)\left[\right(\frac{2r^2{(r+1)}^2}{4})–(\frac{r(r+a)(2r+1)}{6})+(\frac{r(r+1)}{2}\left)\right] \\ =\left(\frac{1}{12}\right)r(r+1)(3r^2+r+2) \end{gathered}$$

When $r=n$

$\sum V_r=\left(\frac{1}{12}\right)n(n+1)(3n2+n+2)$

Hence, the correct option is (A)