Question

Let ${\Delta }_a=\left|\begin{array}{ccc}a-1& n& 6\\ (a-1{)}^2& 2n^2& 4n-2\\ (a-1{)}^3& 3n^3& 3n^2-3n\end{array}\right|$;
the value of ${\sum }_{a=1}^n{\Delta }_a$ is

• A. 1
• B. \N
• C. n
• D. $n^2$

Answer 1

The correct option is B \N

Given, ${\Delta }_a=\left|\begin{array}{ccc}a-1& n& 6\\ (a-1{)}^2& 2n^2& 4n-2\\ (a-1{)}^3& 3n^3& 3n^2-3n\end{array}\right|$
$\therefore {\sum }_{a=1}^n{\Delta }_a=\left|\begin{array}{ccc}{\sum }_{a=1}^n(a-1)& n& 6\\ {\sum }_{a=1}^n(a-1{)}^2& 2n^2& 4n-2\\ {\sum }_{a=1}^n(a-1{)}^3& 3n^3& 3n^2-3n\end{array}\right|$
$\left|\begin{array}{ccc}\frac{n(n-1)}{2}& n& 6\\ \frac{n(n-1)(2n-1)}{6}& 2n^2& 4n-2\\ \frac{n^2(n-1{)}^2}{4}& 3n^3& 3n^2-3n\end{array}\right|=\frac{n^2(n-1)}{2}\left|\begin{array}{ccc}1& 1& 6\\ \frac{(2n-1)}{3}& 2n& 4n-2\\ \frac{n(n-1)}{2}& 3n^2& 3n^2-3n\end{array}\right|=\frac{n^3(n-1)}{12}\left|\begin{array}{ccc}1& 1& 6\\ 2n-1& 6n& 12n-6\\ n-1& 6n& 6n-6\end{array}\right|$
Applying $C_3\to C_3-6C_1$
$=\frac{n^3(n-1)}{12}\left|\begin{array}{ccc}1& 1& 0\\ 2n-1& 6n& 0\\ n-1& 6n& 0\end{array}\right|=0$