Question

If ${△}_r=\left|\begin{array}{ccc}1& n& n\\ 2r& n^2+n+1& n^2+n\\ 2r-1& n^2& n^2+n+1\end{array}\right|$ and $\sum _{r=1}^n{△}_r=56$, then $n$ is equal to

• A. $4$
• B. $6$
• C. $7$
• D. $9$

Answer 1

Answer: (C)

$7$

Putting $r=1,2,3,...,n$ and using the formula
$\sum 1=n$
and $\sum r=\frac{n(n+1)}{2}$
$\sum (2r-1)=1+3+5+....=n^2$
Therefore, $\sum _{r=1}^n{△}_r=\left|\begin{array}{ccc}n& n& n\\ n(n+1)& n^2+n+1& n^2+n\\ n^2& n^2& n^2+n+1\end{array}\right|=56$
Applying $C_1\to C_1-C_3,C_2\to C_2-C_3$, we get
$\left|\begin{array}{ccc}0& 0& n\\ 0& 1& n^2+n\\ -n-1& -n-1& n^2+n+1\end{array}\right|=56$
$\Rightarrow n(n+1)=56$
$\Rightarrow n^2+n-56=0$
$\Rightarrow (n+8)(n-7)=0$
$\Rightarrow n=7,(n\ne -8)$.