# If $S_r=\begin{vmatrix}2r&x&n(n+1)\\6r^2-1&y&n^2(2n+3)\\4r^32nr&z&n^3(n+1)\end{vmatrix}$ then value of $\sum\limits_{r=1}^n S_r$ is independent of

$\begin{array}{1 1}(A)\;\text{x only}&(B)\;\text{y only}\\(C)\;\text{x,y,z,n }&(D)\;\text{n only}\end{array}$

## 1 Answer

$\sum\limits_{r=1}^n S_r=\begin{vmatrix} \sum\limits_{r=1}^n 2r& x&n(n+1)\\\sum\limits_{r=1}^n (6r^2-1)& y&n^2(2n+3)\\\sum\limits_{r=1}^n (4r^2-2nr)& z&n^3(n+1)\end{vmatrix}$
$\Rightarrow \begin{vmatrix} n(n+1)&x&n(n+1)\\n(n+1)(2n+1)-n&y&n^2(2n+3)\\n^2(n+1)^2-n^2(n+1)& z&n^3(n+1)\end{vmatrix}$
$\Rightarrow \begin{vmatrix} n(n+1)&x&n(n+1)\\n^2(2n+3)&y&n^2(2n+3)\\n^3(n+1)& z&n^3(n+1)\end{vmatrix}=0$
And hence independent of $x,y,z,n$
Hence (C) is the correct answer.
answered Apr 22, 2014

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer