Question

If $\left|\begin{array}{ccc}1& n& n\\ 2r& n^2+n+1& n^2+n\\ 2r+1& n^2& n^2+n+1\end{array}\right|$ and ${\sum }_{r=1}^n{\Delta }_r=110$.
Then value of $n$ equals to,

• A. $11$
• B. $-10$
• C. $10$
• D. None of these

Answer 1

Answer: (A)

$11$

Given, ${\Delta }_r=\left|\begin{array}{ccc}1& n& n\\ 2r& n^2+n+1& n^2+n\\ 2r+1& n^2& n^2+n+1\end{array}\right|$
${\sum }_{r=1}^n{\Delta }_r=\left|\begin{array}{ccc}{\sum }_{r=1}^{n}1& n& n\\ 2{\sum }_{r=1}^{n}r& n^2+n+1& n^2+n\\ {\sum }_{r=1}^n(2r+1)& n^2& n^2+n+1\end{array}\right|$
$110=\left|\begin{array}{ccc}n& n& n\\ n(n+1)& n^2+n+1& n^2+n\\ n(n+1)+n& n^2& n^2+n+1\end{array}\right|$
$110=n\left|\begin{array}{ccc}1& 1& 1\\ n^2+n& n^2+n+1& n^2+n\\ n^2+n+n& n^2& n^2+n+1\end{array}\right|$
$C_1\to C_1-C_3$
$110=n(-1)\left|\begin{array}{ccc}0& 1& 1\\ 0& n^2+n+1& n^2+n\\ n-1& n^2& n^2+n+1\end{array}\right|$
$\Rightarrow 11\times 10=n(n-1)$
$\Rightarrow n=11$