Question

If $D_r=$ $\left|\begin{array}{ccc}r-1& n& 6\\ {(r-1)}^2& 2n^2& 4n-2\\ {(r-1)}^3& 3n^3& 3n^2-3n\end{array}\right|$ then $\sum _{r=1}^n{D}_r=$

• A. $nr$
• B. $0$
• C. $\frac{n(n-1)}{2}-r^2$
• D. $2n-n^2$

Answer 1

The correct option is B $0$

$D_r=\left|\begin{array}{ccc}r-1& n& 6\\ {(r-1)}^2& 2n^2& 4n-2\\ {(r-1)}^3& 3n^3& 3n^2-3n\end{array}\right|$
$\Rightarrow {\sum }_{r=1}^n{D}_r=\left|\begin{array}{ccc}0& 2\left(3^0\right)& 4\left(5^0\right)\\ 0& y& z\\ 0& 3^n-1& 5^n-1\end{array}\right|+\left|\begin{array}{ccc}1& n& 6\\ 1^2& 2n^2& 4n-2\\ 1^3& 3n^3-1& 3n^2-3n\end{array}\right|+\left|\begin{array}{ccc}2& n& 6\\ 2^2& 2n^2& 4n-2\\ 2^3& 3n^3-1& 3n^2-3n\end{array}\right|+.......\left|\begin{array}{ccc}n-1& n& 6\\ {(n-1)}^2& 2n^2& 4n-2\\ (n-1{)}^3& 3n^3-1& 3n^2-3n\end{array}\right|$
$=\left|\begin{array}{ccc}1+2+3+....+(n-1)& n& 6\\ {1^2+2^2+3^2+....+(n-1)}^2& 2n^2& 4n-2\\ 1^3+2^3+3^3+....(n-1{)}^3& 3n^3-1& 3n^2-3n\end{array}\right|$
$=\left|\begin{array}{ccc}\frac{n(n-1)}{2}& n& 6\\ \frac{n(n-1)(2n-1)}{6}& 2n^2& 4n-2\\ \frac{n^2(n-1{)}^2}{4}& 3n^3-1& 3n^2-3n\end{array}\right|$
$=\frac{n(n-1)}{2}\left|\begin{array}{ccc}1& n& 6\\ \frac{(2n-1)}{3}& 2n^2& 4n-2\\ \frac{n(n-1)}{2}& 3n^3-1& 3n^2-3n\end{array}\right|$
$=\frac{n(n-1)}{12}\left|\begin{array}{ccc}1& n& 6\\ (2n-1)& 6n^2& 12n-6\\ n(n-1)& 6n^3-2& 6n^2-6n\end{array}\right|$
$=\frac{6n(n-1)}{12}\left|\begin{array}{ccc}1& n& 1\\ (2n-1)& 6n^2& (2n-1)\\ n(n-1)& 6n^3-2& (n^2-n)\end{array}\right|$
$=0$