Question

If ${\Delta }_r=\left|\begin{array}{ccc}r& n& 6\\ r^2& {2n}^2& 4n-2\\ r^3& {3n}^3& 3n^2-3n\end{array}\right|$, then $\sum _{r=0}^{n-1}{\Delta }_r$ equals to

• A. $0$
• B. $n{(n+2)}^2$
• C. $\frac{1}{12}n(n^3+2)$
• D. None of these

Answer 1

The correct option is A $0$

Let, ${\Delta }_r=\left|\begin{array}{ccc}r& n& 6\\ r^2& 2n^2& 4n-2\\ r^3& 3n^3& 3n^2-3n\end{array}\right|$,

then $\sum _{r=0}^{n-1}{\Delta }_r=$

$\left|\begin{array}{ccc}\sum _{r=0}^{n-1}r& n& 6\\ {\sum }_{r=0}^{n-1}{r}^2& 2n^2& 4n-2\\ \sum _{r=0}^{n-1}{r}^3& 3n^3& 3n^2-3n\end{array}\right|$

$=\left|\begin{array}{ccc}\frac{n(n-1)}{2}& n& 6\\ \frac{n(n-1)(2n-1)}{6}& 2n^2& 4n-2\\ {\left(\frac{n(n-1)}{2}\right)}^2& 3n^3& 3n^2-3n\end{array}\right|$

$=\frac{n(n-1)}{2}\left|\begin{array}{ccc}1& n& 6\\ \frac{(2n-1)}{3}& 2n^2& 4n-2\\ \frac{n(n-1)}{2}& 3n^3& 3n^2-3n\end{array}\right|$

$=\frac{n(n-1)}{12}\left|\begin{array}{ccc}1& n& 6\\ (2n-1)& 6n^2& 12n-6\\ n(n-1)& 6n^3& 6n^2-6n\end{array}\right|$

[$C_3^{\prime }=C_3-6C_1$]

$=\frac{n(n-1)}{12}\left|\begin{array}{ccc}1& n& 0\\ (2n-1)& 6n^2& 0\\ n(n-1)& 6n^3& 0\end{array}\right|=0$