If $f(n)=\begin{vmatrix} n& 1&5\\n^2&2r+1&2r+1\\n^3&3r^2&3r+1\end{vmatrix}$ then $\sum\limits_{n=1}^ r f(n)$ is

$\begin{array}{1 1}(A)\;2\sum \limits_{n=1}^r n&(B)\;2\sum\limits_{n=1}^r n^2\\(C)\;\large\frac{1}{2}\normalsize \sum \limits_{n=1}^r n^2&(D)\;0\end{array}$

$\sum\limits_{n=1}^ r f(n)=\begin{vmatrix} \sum\limits_{n=1}^r n& 1&5\\\sum\limits_{n=1}^r n^2&2r+1&2r+1\\\sum\limits_{n=1}^rn^3&3r^2&3r+1\end{vmatrix}$
$\Rightarrow \begin{vmatrix} \large\frac{r(r+1)}{2}&1&5\\\large\frac{r(r+1)(2r+1)}{6}&2r+1&2r+1\\\large\frac{r^2(r+1)^2}{4}&3r^2&3r+1\end{vmatrix}$
$\Rightarrow \large\frac{1}{12}$$r(r+1)\begin{vmatrix} 6& 1& 5\\2(2r+1)& 2r+1&2r+1\\3r(r+1)&3r^2&3r+1\end{vmatrix} Apply C_1=C_1-C_2-C_3 \Rightarrow \large\frac{1}{12}$$r(r+1)\begin{vmatrix}0& 1&5\\0&2r+1& 2r+1\\-1&3r^2&3r+1\end{vmatrix}$
$\Rightarrow \large\frac{1}{12}$$r(r+1)(-1)[2r+1-5(2r+1)] \Rightarrow \large\frac{1}{3}$$r(r+1)(2r+1)$
$\Rightarrow 2\sum\limits_{n=1}^ r n^2$
Hence (B) is the correct answer.