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

$D_k=\left|\begin{array}{ccc}1& n& n\\ 2k& n^2+n+1& n^2+n\\ 2k-1& n^2& n^2+n+1\end{array}\right|$

${\sum }_{k=1}^n{D}_k=\left|\begin{array}{ccc}{\sum }_{k=1}^{n}1& n& n\\ {\sum }_{k=1}^{n}2k& n^2+n+1& n^2+n\\ {\sum }_{k=1}^n(2k-1)& n^2& n^2+n+1\end{array}\right|56=\left|\begin{array}{ccc}n+1& n& n\\ 2\frac{n(n+1)}{2}& n^2+n+1& n^2+n\\ 2\frac{n(n+1)}{2}-(n+1)& n^2& n^2+n+1\end{array}\right|$

$56=\left|\begin{array}{ccc}n+1& n& n\\ n^2+n& n^2+n+1& n^2+n\\ n^2-1& n^2& n^2+n+1\end{array}\right|$

Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$
$56=\left|\begin{array}{ccc}1& -1& -1\\ n^2+n& 1& 0\\ n^2-1& 1& n+2\end{array}\right|$
Applying $R_2\to R_2+R_1,R_3\to R_3+R_1$
$56=\left|\begin{array}{ccc}1& -1& -1\\ n^2+n+1& 0& -1\\ n^2& 0& n+1\end{array}\right|56=-1((n+1)(n^2+n+1)+n^2)56=-1(n^3+n^2+n+n^2+n+1+n^2)n^3+3n^2+2n=-57$