Question

If the value of determinant $\left|\begin{array}{ccc}{n}^2& (n+1{)}^2& (n+2{)}^2\\ (n+1{)}^2& (n+2{)}^2& (n+3{)}^2\\ (n+2{)}^2& (n+3{)}^2& (n+4{)}^2\end{array}\right|=p\forall n\in N$ and $p\in R.$ Then the value of $p$ is

• A. $8$
• B. $-8$
• C. $16$
• D. $-25$

Answer 1

The correct option is B $-8$

Given : $\Delta =\left|\begin{array}{ccc}{n}^2& (n+1{)}^2& (n+2{)}^2\\ (n+1{)}^2& (n+2{)}^2& (n+3{)}^2\\ (n+2{)}^2& (n+3{)}^2& (n+4{)}^2\end{array}\right|$
(as, the above determinant is true for all $n\in N$)
put $n=1$ in the above determinant,
$\Rightarrow \Delta =\left|\begin{array}{ccc}1& 4& 9\\ 4& 9& 16\\ 9& 16& 25\end{array}\right|$
expanding along $R_1,$
$\Rightarrow \Delta =1(225-256)-4(100-144)+9(64-81)=-8$
$\therefore p=-8$