If $D_r=\begin{vmatrix}r & 1 & \frac{n(n+1)}{2}\\2r-1 & 4 & n^2\\2^{r-1} & 5 & 2^n-1\end{vmatrix},$ then the value of $\sum\limits^{n}_{r=1}D_r$, is
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : 0
We have,
$\sum\limits^{n}_{r=1}D_r=\begin{vmatrix}\sum\limits^{n}_{r=1}r& 1 & \frac{n(n+1)}{2}\\\sum\limits^{n}_{r=1}2r-1 & 4 & n^2\\\sum\limits^{n}_{r=1}2^{r-1} & 5 & 2^n-1\end{vmatrix}$
$\sum\limits^{n}_{r=1}D_r=\begin{vmatrix}\frac{n(n+1)}{2} & 1 & \frac{n(n+1)}{2}\\n^2 & 4 & n^2\\2^{r-1} & 5 & 2^n-1\end{vmatrix}=0$