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 $\sum\limits_{r=1}^{n}D_r$, is
Answer & explanation
Correct answer: option 1
We have,
$\sum\limits_{r=1}^{n}D_r=\begin{vmatrix}\sum\limits_{r=1}^{n}r&1&\frac{n (n + 1)}{2}\\\sum\limits_{r=1}^{n}2r-1 &4&n^2\\\sum\limits_{r=1}^{n}2^{r-1}&5&2^n-1\end{vmatrix}$
$⇒\sum\limits_{r=1}^{n}D_r=\begin{vmatrix}\frac{n (n + 1)}{2}&1&\frac{n (n + 1)}{2}\\n^2 &4&n^2\\2^n-1&5&2^n-1\end{vmatrix}=0$