Question

If ${\Delta }_r=\left|\begin{array}{ccc}1& n& n\\ 2r& n^2+n+1& n^2+n\\ 2r-1& n^2& n^2+n+1\end{array}\right|$ and ${\sum }_{r=1}^n{\Delta }_r=56$ then n equals

• A. 4
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (C)

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)
${\sum }_{r=1}^n{\Delta }_r=\left|\begin{array}{ccc}n& n& n\\ n\left(n+1\right)& n^2+n+1& n^2+n\\ n^2& n^2& n^2+n+1\end{array}\right|=56$
Applying $C_1\to C_1-C_3,C_2\to C_2-C_3$
$\left|\begin{array}{ccc}0& 0& n\\ 0& 1& n^2+n\\ -n-1& -n-1& n^2+n+1\end{array}\right|=56$
or n (n+1) = 56 or n2 + n - 56 = 0
(n + 8) (n - 7) = 0 $\Rightarrow$ n = 7 (-8 is rejected).