Thank you for reporting, we will resolve it shortly
Q.
If $\Delta_{r} = \begin{vmatrix}1&n&n\\ 2r&n^{2} +n+1&n^{2}+n\\ 2r-1&n^{2}&n^{2}+n+1\end{vmatrix} $ and $\sum^{n}_{r=1} \Delta_{r} = 56 $ then n equals
Determinants
Solution:
Putting r = 1, 2, 3, .....n and using the formula
$\sum 1 = n , 2 \sum r = \frac{2(n + 1)n}{2}$
$\sum(2r - 1 ) = 1 + 3 + 5+ .... = n^2$ (sum of A.P)
$\sum ^{n}_{r=1} \Delta _{r} = \begin{vmatrix}n&n&n\\ n\left(n+1\right)&n^{2}+n+1&n^{2}+n\\ n^{2}&n^{2}&n^{2}+n+1\end{vmatrix}=56 $
Applying $C_{1} \to C_{1} - C_{3}, C_{2} \to C_{2} -C_{3}$
$\begin{vmatrix}0&0&n\\ 0&1&n^{2}+n\\ -n-1&-n-1&n^{2}+n+1\end{vmatrix} = 56 $
or n (n+1) = 56 or n2 + n - 56 = 0
(n + 8) (n - 7) = 0 $\Rightarrow $ n = 7 (-8 is rejected).