$V_r=$ Sum of first $r$ terms of an $\mathrm{AP}$ whose first term is $r$ and common difference is $\left(2r-1\right)$

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

$=\frac{r}{2}\left[2r+2r^2-3r+1\right]$

$=\frac{r}{2}\left[2r^2-r+1\right]$

$=\frac{1}{2}\left[2r^3-r^2+r\right]$

Now, $\sum _{r=1}^n{V}_r=\frac{1}{2}\left[2\sum _{r=1}^n{r}^3-\sum _{r=1}^n{r}^2+\sum _{r=1}^{n}r\right]$

$=\frac{1}{2}\left[2·{\left(\frac{n\left(n+1\right)}{2}\right)}^2-\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right]$

$=\frac{1}{2}\left[\frac{n^2{\left(n+1\right)}^2}{2}-\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right]$

$=\frac{n\left(n+1\right)}{4}\left[n\left(n+1\right)-\frac{\left(2n+1\right)}{3}+1\right]$

$=\frac{n\left(n+1\right)}{4}\left[\frac{3n^2+3n-2n-1+3}{3}\right]$

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