# If $\Delta_r=\begin{vmatrix}2^{r-1}&\large\frac{(r+1)!}{1+\large\frac{1}{r}}&2r\\a&b&c\\2^n-1&(n+1)!-1&n(n+1)\end{vmatrix}$ then value of $\sum\limits_{r=1}^n\Delta_r$ is

$\begin{array}{1 1}(A)\;0&(B)\;(n+3)!\\(C)\;a(n!)+b&(D)\;\text{None of these}\end{array}$

$\sum\limits_{r=1}^n 2^{r-1}=2^n-1\sum\limits_{r=1}^n r=n\big[\large\frac{n(n+1)}{2}\big]$$=n(n+1) \sum\limits_{r=1}^n\large\frac{(r+1)!}{1+\large\frac{1}{r}}$$=\sum\limits_{r=1}^n r(r!)=\sum\limits_{r=1}^n (r+1-1)(r!)$
$\Rightarrow \sum_{r=1}^n\big[(r+1)!-r!\big]=(n+1)!-1$
$\therefore \sum_{r=1}^n \Delta_r=\begin{vmatrix}\sum\limits_{r=1}^n 2^{r-1}&\sum _{r=1}^n\large\frac{(r+1)!}{1+\large\frac{1}{r}}&2\sum_{r=1}^n r\\a&b&c\\2^n-1&(n+1)!-1&n(n+1)\end{vmatrix}=0$
($R_1$ and $R_3$ are identical)
Hence (A) is the correct answer.