Question

If $D_k=\left|\begin{array}{ccc}1& n& n\\ 2k& n^2+n+1& n^2+n\\ 2k-1& n^2& n^2+n+1\end{array}\right|$ and $\sum _{k=1}^n{D}_k=56$ then n equals

• A. 4
• B. 6
• C. 8
• D. none of these

Answer 1

The correct option is D none of these

$D_k=\left|\begin{array}{ccc}1& n& n\\ 2k& n^2+n+1& n^2+n\\ 2k-1& n^2& n^2+n+1\end{array}\right|$
${\sum }_{k=1}^n{D}_k=\left|\begin{array}{ccc}{\sum }_{k=1}^{n}1& n& n\\ 2{\sum }_{k=1}^{n}k& n^2+n+1& n^2+n\\ 2{\sum }_{k=1}^{n}k-1& n^2& n^2+n+1\end{array}\right|$
$\Rightarrow 56=\left|\begin{array}{ccc}n& n& n\\ n(n+1)& n^2+n+1& n^2+n\\ n^2+n-1& n^2& n^2+n+1\end{array}\right|$
Applying $C_1\to C_1-C_2$ and $C_2\to C_2-C_3$
$\Rightarrow 56=n\left|\begin{array}{ccc}0& 0& 1\\ -1& 1& n^2+n\\ n-1& -(n+1)& n^2+n+1\end{array}\right|$
$\Rightarrow 2n=56$
$\Rightarrow n=28$