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 $\sum _{k=1}^n{D}_k=56,$ where $n\in N,$ then $n$ is equal to

Answer 1

It is given that $\sum _{k=1}^n{D}_k=56$
where, $D_k=\left|\begin{array}{ccc}1& n& n\\ 2k& n^2+n+1& n^2+n\\ 2k-1& n^2& n^2+n+1\end{array}\right|$

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

$\sum _{k=1}^{n}1=n$
$\sum _{k=1}^{n}2k=2(1+2+\cdots +n)=\frac{2n(n+1)}{2}=n(n+1)$
$\sum _{k=1}^n(2k-1)=\sum _{k=1}^{n}2k-\sum _{k=1}^{n}1=n(n+1)-n=n^2$

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

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

$\Rightarrow n(n+1)=56$
$\Rightarrow n^2+n-56=0$
$\Rightarrow n^2+8n-7n-56=0$
$\Rightarrow (n+8)(n-7)=0$
$\Rightarrow n=7,n=-8$ (Not Possible)
$\therefore n=7$