If $Δ_r=\begin{vmatrix}2^{r-1}&\frac{(r + 1)!}{(1 + 1/r)}&2r\\a&b&c\\2^n-1 &(n+1)!-1&n(n+1)\end{vmatrix}$, then $\sum\limits_{r=1}^{n}Δ_r$ is equal to

0

We have,

$Δ_r=\begin{vmatrix}2^{r-1}&r(r!)&2r\\a&b&c\\2^n-1 &(n+1)!-1&n(n+1)\end{vmatrix}$

$∴\sum\limits_{r=1}^{n}Δ_r=\begin{vmatrix}\sum\limits_{r=1}^{n}2^{r-1}&\sum\limits_{r=1}^{n}r(r!)&\sum\limits_{r=1}^{n}2r\\a&b&c\\2^n-1 &(n+1)!-1&n(n+1)\end{vmatrix}$

But, $r (r!)=(r+1)!-r!$

$∴\sum\limits_{r=1}^{n}r (r!)=\sum\limits_{r=1}^{n}[(r+1)! -r!]=(n+1)!-1$

$∴\sum\limits_{r=1}^{n}Δ_r=\begin{vmatrix}2^n-1 &(n+1)!-1&n(n+1)\\a&b&c\\2^n-1 &(n+1)!-1&n(n+1)\end{vmatrix}$

$⇒\sum\limits_{r=1}^{n}Δ_r=0$  [∵ $R_1$ and $R_3$ are identical]