# Solve the following equations

Question:

If $D_{k}=\left|\begin{array}{ccc}1 & n & n \\ 2 k & n^{2}+n+2 & n^{2}+n \\ 2 k-1 & n^{2} & n^{2}+n+2\end{array}\right|$ and $\sum_{k=1}^{n} D_{k}=48$, then $n$ equals

(a) 4

(b) 6

(c) 8

(d) none of these

Solution:

(a) 4

$D_{k}=\mid \begin{array}{lll}1 & n & n\end{array}$

$2 k \quad n^{2}+n+2 \quad n^{2}+n$

$\begin{array}{lll}2 k-1 & n^{2} & n^{2}+n+2 \mid\end{array}$

$=\mid 1 \quad n \quad n$

$\begin{array}{lll}1 & n+2 & -2\end{array}$

$\begin{array}{lll}2 k-1 & n^{2} & n^{2}+n+2 \mid\end{array}$       $\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{3}\right]$

$=\mid 1 \quad n \quad n$

$\begin{array}{lll}0 & 2 & -2-n\end{array}$

$2 k-1 \quad n^{2} \quad n^{2}+n+2 \mid$                $\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\right]$

$\begin{array}{lll}0 & 2 & -2-n\end{array}$

$\begin{array}{llllllll}1 & n^{2} & n^{2}+n+2|+| 1 & n & n 0 & 2 & -2-n 3\end{array} n^{2}+n+2 \mid+\ldots$

$\begin{array}{ccc}+\mid 1 & n & n \\ 0 & 2 & -2-n \\ n & n^{2} & n^{2}+n+2 \mid\end{array}$

$\sum_{k=1}^{n} D_{k}=1\left(2\left(n^{2}+n+2\right)+(2+n) n^{2}\right)+1(n(-2-n)-2 n)+1\left(2\left(n^{2}+n+2\right)+(2+n) n^{2}\right)+2(n(-2-n)-2 n)+\ldots$

$+1\left(2\left(n^{2}+n+2\right)+(2+n) n^{2}\right)+n(n(-2-n)-2 n)$

$\sum_{k=1}^{n} D_{k}=n\left(2\left(n^{2}+n+2\right)+(2+n) n^{2}\right)+(n(-2-n)-2 n)(1+3+5+7+\ldots+n)$

$\sum_{k=1}^{n} D_{k}=n\left(2\left(n^{2}+n+2\right)+(2+n) n^{2}\right)+(n(-2-n)-2 n)\left(n^{2}\right)$

$\sum_{k=1}^{n} D_{k}=2 n^{2}+4 n$

$\Rightarrow 2 n^{2}+4 n=48$

$\Rightarrow(n-6)(n-4)=0$

$\Rightarrow n=4$