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

$\therefore {\sum }_{k=1}^n{D}_k=\left|\begin{array}{ccc}{\sum }_{k=1}^{n}1& n& n\\ 2{\sum }_{k=1}^{n}k& n^2+n+1& n^2+n\\ 2{\sum }_{k=1}^{n}k-{\sum }_{k=1}^{n}1& n^2& n^2+n+1\end{array}\right|$
$=\left|\begin{array}{ccc}n& n& n\\ n^2+n& n^2+n+1& n^2+n\\ n^2& n^2& n^2+n+1\end{array}\right|=56$
=Applying $C_2\to C_2\to C_1\to C_3\to C_3-C_1$, then
$\left|\begin{array}{ccc}n& 0& 0\\ n^2+n& 1& 0\\ n& 0& n+1\end{array}\right|=56\Rightarrow n(n+1)=56=7\times 8\Rightarrow n=7$