Question

If $D_r=\left|\begin{array}{ccc}r& x& \frac{n(n+1)}{2}\\ 2r-1& y& n^2\\ 3r-2& z& \frac{n(3n-1)}{2}\end{array}\right|$, then $\sum _{r=1}^n{D}_r$ is equal to

• A. $\frac{1}{6}n(n+1)(2n+1)$
• B. $\frac{1}{4}{n}^2(n+1{)}^2$
• C. $0$
• D. none of these

Answer 1

Answer: (C)

$0$

$Dr=\left[\begin{array}{ccc}r& x& \frac{n(n+1)}{2}\\ 2r-1& y& n^2\\ 3r-2& z& \frac{n(3n-1)}{2}\end{array}\right]$

${\sum }_{r=1}^{n}Dr=\left|\begin{array}{ccc}\frac{n(n+1)}{2}& x& \frac{n(n+1)}{2}\\ 2\left(\frac{n(n+1)}{2}\right)-n& y& n^2\\ 3\left(\frac{n(n+1)}{2}\right)-2n& z& n\frac{(3n-1)}{2}\end{array}\right|$

$=\left|\begin{array}{ccc}\frac{n(n+1)}{2}& x& \frac{n(n+1)}{2}\\ n^2& y& n^2\\ \frac{n(3n-1)}{2}& z& \frac{n(3n-1)}{2}\end{array}\right|=0$