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 $\sum _{r=1}^n{S}_r$ does not depend on

• A. $x$
• B. $y$
• C. $n$
• D. all of these

Answer 1

The correct option is D all of these

${\sum }_{r=1}^{n}sr=\left|\begin{array}{ccc}2\sum r& x& n(n+1)\\ 6\sum r^2-\sum r& y& n^2(2n+3)\\ 4\sum r^3-2n\sum r& z& n(n+1)\end{array}\right|$

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

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

Two columns are same ${\sum }_{r=1}^{n}sr=0$