Question

If $D_k=\left|\begin{array}{ccc}1& n& n\\ 2k& n^2+n+2& n^2+n\\ 2k-1& n^2& n^2+n+2\end{array}\right|$ , $\sum _{k=1}^n{D}_k=48$, then n equals

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

Answer 1

The correct option is A 4

$D_k=\left|\begin{array}{ccc}1& n& n\\ 2k& n^2+n+2& n^2+n\\ 2k-1& n^2& n^2+n+2\end{array}\right|$
$\sum _{k=1}^n{D}_k=48$
$\to \left|\begin{array}{ccc}{\sum }_{k=1}^{n}1& n& n\\ {\sum }_{k=1}^{n}2k& n^2+n+2& n^2+n\\ {\sum }_{k=1}^{n}2k-1& n^2& n^2+n+2\end{array}\right|=48$
$\to \left|\begin{array}{ccc}n& n& n\\ n^2+n& n^2+n+2& n^2+n\\ n^2& n^2& n^2+n+2\end{array}\right|=48$
applying $C_3\to C_3-C_1$ and $C_2\to C_2-C_1$
$\to \left|\begin{array}{ccc}n& 0& 0\\ n^2+n& 2& 0\\ n^2& 0& n+2\end{array}\right|=48$
$\Rightarrow n(n+2)=24$
$\therefore n=4$