Question

The sum $V_1+V_2+V_3+...+V_n$ is

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

Answer 1

The correct option is A $\frac{n(n+1)(3n^2+n+2)}{12}$

We have sum of $n$ terms$=\frac{n}{2}(2a+(n-1\left)d\right)$ where $a$ is the first term, $n$ is the number of terms and $d$ is the common difference in an A.P.
From the passage,$n=r$, $a=2r$ and $d=(2r-1)$
$\therefore V_r=\frac{r}{2}[2r+(r-1\left)\right(2r-1\left)\right]$
$\Rightarrow \frac{r}{2}[2r+2r^2-3r+1]=\frac{r}{2}[2r^2-r+1]$
Thus,$V_r=\frac{1}{2}[2r^3-r^2+r]=r^3-\frac{r^2}{2}+\frac{r}{2}$
Now,$V_1+V_2+V_3+...+V_n={\sum }_{r=1}^n{V}_r$
${\sum }_{r=1}^n{V}_r={\sum }_{r=1}^n(r^3-\frac{r^2}{2}+\frac{r}{2})$
Splitting all the above terms we get
$={\sum }_{r=1}^n{r}^3-{\sum }_{r=1}^n\frac{r^2}{2}+{\sum }_{r=1}^n\frac{r}{2}$
Substituting the formula of summation of series
$={\left(\frac{n(n+1)}{2}\right)}^2-\frac{1}{2}\frac{n(n+1)(2n+1)}{6}+\frac{1}{2}\frac{n(n+1)}{2}$
Taking common term,
$\frac{n(n+1)}{2}[\frac{n(n+1)}{2}-\frac{2n+1}{6}+\frac{1}{2}]$
$\frac{n(n+1)}{2}(3n^2+3n-2n-1+3)$
$=\frac{n(n+1)}{12}(3n^2+n+2)$