#### Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 8 maths

Correct option (a)

Hint:

Solve the given determinant Dk

Given:

$D_{k}=\begin{vmatrix} 1 &n &n \\ 2k &n^{2}+n+2 &n^{2}+n \\ 2k-1 &n^{2} &n^{2}+n+2 \end{vmatrix}\; \; and\; \; \sum_{k=1}^{n}D_{k}=48$

We have to find the value of n.

Solution:

$D_{k}=\begin{vmatrix} 1 &n &n \\ 2k &n^{2}+n+2 &n^{2}+n \\ 2k-1 &n^{2} &n^{2}+n+2 \end{vmatrix}$

Applying row transformation,

R2→R2-R3, We get

$D_{k}=\begin{vmatrix} 1 &n &n \\ 1 &n+2 &-2 \\ 2k-1 &n^{2} &n^{2}+n+2 \end{vmatrix}$

Applying R2→R2-R3, We get

$D_{k}=\begin{vmatrix} 0 &-2 &n+2 \\ 1 &n+2 &-2 \\ 2k-1 &n^{2} &n^{2}+n+2 \end{vmatrix}$

Expanding along Row1, we get

$=2(n^{2}+n+2+4k-2)+(n+2)(n^{2}-(n-2)(2k-1))$

$=2(n^{2}+n+4k)+(n+2)(n^{2}-2nk+n-4k+2)$

$=2n^{2}+2n+8k+n^{3}-2n^{2}k+n^{2}-4nk+2n+2n^{2}-4nk+2n-8k+4$

$D_{k}=n^{3}+5n^{2}-2n^{2}k+6n-8nk+4$

Given:

$\sum_{k=1}^{n}D_{k}=48$

$\Rightarrow \sum_{k=1}^{n}1(n^{3}+5n^{2}+6n+4)-2n^{2}\sum_{k=1}^{n}k-8n\sum_{k=1}^{n}k=48$

$\Rightarrow n(n^{3}+5n^{2}+6n+4)-2n^{2}\frac{n(n+1)}{2}-8n\frac{n(n+1)}{2}=48$

$\Rightarrow n^{4}+5n^{3}+6n^{2}+4n-n^{4}-n^{3}-4n^{3}-4n^{2}=48$

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

$\Rightarrow n^{2}+2n-24=0$

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

$\Rightarrow n=-6,4$

Hence according to option n=4