Question

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

(a) 4
(b) 6
(c) 8
(d) none of these

Answer 1

(a) 4

$$\begin{gathered} D_k=\left|1nn \\ 2k{n}^2+n+2n^2+n \\ 2k-1n^2{n}^2+n+2\right| \\ =\left|1nn \\ 1n+2-2 \\ 2k-1n^2{n}^2+n+2\right|\left[\mathrm{Applying}{R}_2\to R_2-R_3\right] \\ =\left|1nn \\ 02-2-n \\ 2k-1n^2{n}^2+n+2\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\right] \\ {\sum }_{k=1}^n{D}_k=\left|1nn \\ 02-2-n \\ 1n^2{n}^2+n+2\right|+\left|1nn \\ 02-2-n \\ 3n^2{n}^2+n+2\right|+...+\left|1nn \\ 02-2-n \\ n{n}^2{n}^2+n+2\right| \\ {\sum }_{k=1}^n{D}_k=1\left(2\left(n^2+n+2\right)+\left(2+n\right)n^2\right)+1\left(n\left(-2-n\right)-2n\right)+1\left(2\left(n^2+n+2\right)+\left(2+n\right)n^2\right)+2\left(n\left(-2-n\right)-2n\right)+...+1\left(2\left(n^2+n+2\right)+\left(2+n\right)n^2\right)+n\left(n\left(-2-n\right)-2n\right) \\ {\sum }_{k=1}^n{D}_k=n\left(2\left(n^2+n+2\right)+\left(2+n\right)n^2\right)+\left(n\left(-2-n\right)-2n\right)\left(1+3+5+7+...+n\right) \\ {\sum }_{k=1}^n{D}_k=n\left(2\left(n^2+n+2\right)+\left(2+n\right)n^2\right)+\left(n\left(-2-n\right)-2n\right)\left(n^2\right) \\ {\sum }_{k=1}^n{D}_k=2n^2+4n \\ \Rightarrow 2n^2+4n=48 \\ \Rightarrow \left(n-6\right)\left(n-4\right)=0 \\ \Rightarrow n=4 \end{gathered}$$