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 $Q_r=T_{r+1}-T_r$ for $r=1,2,...$

The sum $V_1+V_2+...+V_n$ is

• A. $\frac{1}{12}n(n+1)(3n^2-n+1)$
• B. $\frac{1}{12}n(n+1)(3n^2+n+2)$
• C. $\frac{1}{2}n(2n^2-n+1)$
• D. $\frac{1}{3}(2n^3-2n+3)$

Answer 1

Answer: (B)

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

$V_r=\frac{r}{2}(r+(r-1\left)\right(2r-1\left)\right)$

$=\frac{1}{2}(2r^3-r^2+r)$

$V_1+V_2+V_3+.......=\sum V_r$

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

$=\frac{1}{2}(2\times \frac{r^2(r+1{)}^2}{4}-\frac{r(r+a)(2r+1)}{6}+\frac{r(r+1)}{2})$

$=\frac{1}{12}r(r+1)(3r^2+r+2)$

for $r=n$ (i.e. sum upto n terms)

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