Question

If $S_r=\left|\begin{array}{ccc}2r& x& n(n+1)\\ 6r^2-1& y& n^2(2n+3)\\ 4r^3-2nr& z& n^3(n+1)\end{array}\right|$, then the value of ${\sum }_{r=1}^n{S}_r$ is independent of

• A. $x$ only
• B. $y$ only
• C. $n$ only
• D. $x,y,z$ and $n$

Answer 1

The correct option is D $x,y,z$ and $n$

${\sum }_{r=1}^n{S}_r$= $\left|\begin{array}{ccc}{\sum }_{r=1}^{n}2r& x& n(n+1)\\ {\sum }_{r=1}^{n}6r^2-1& y& n^2(2n+3)\\ {\sum }_{r=1}^{n}4r^3-2nr& z& n^3(n+1)\end{array}\right|$
Now, ${\sum }_{r=1}^{n}2r=\frac{2\left(n\right)(n+1)}{2}=n(n+1)$
and ${\sum }_{r=1}^{n}6r^2-1={\sum }_{r=1}^{n}6r^2-{\sum }_{r=1}^{n}1=\frac{6\left(n\right)(n+1)(2n+1)}{6}-n=2n^3+3n^2=n^2(2n+3)$
and ${\sum }_{r=1}^{n}4r^3-2nr={\sum }_{r=1}^{n}4r^3-{\sum }_{r=1}^{n}2nr=\frac{4\left(n\right(n+1){)}^2}{4}-\frac{2n^2(n+1)}{2}=n^3(n+1)$
So we get
$\left|\begin{array}{ccc}{\sum }_{r=1}^{n}2r& x& n(n+1)\\ {\sum }_{r=1}^{n}6r^2-1& y& n^2(2n+3)\\ {\sum }_{r=1}^{n}4r^3-2nr& z& n^3(n+1)\end{array}\right|=$ $\left|\begin{array}{ccc}n(n+1)& x& n(n+1)\\ n^2(2n+3)& y& n^2(2n+3)\\ n^3(n+1)& z& n^3(n+1)\end{array}\right|=0$
Therefore answer is independent of x,y,z and n.
so $D$ is correct