Question

If S_r = $\left| \begin{matrix} 2r & x & n(n + 1) \\ 6r^{2}–1 & y & n^{2}(2n + 3) \\ 4r^{3}–2nr & z & n^{3}(n + 1) \end{matrix} \right|$, then value of $\sum_{r = 1}^{n}{\mspace{6mu} S_{r}}$ is

independent of –

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

Answer 1

Answer: (C)

x, y, z, n

Answer 2

$\sum_{r = 1}^{n}{S_{r} = \mspace{6mu}}\left| \begin{matrix} 2\Sigma r & x & n(n + 1) \\ 6\Sigma r^{2} - \Sigma 1 & y & n^{2}(2n + 3) \\ 4\Sigma r^{3} - 2n\Sigma r & z & n^{3}(n + 1) \end{matrix} \right|$

\frac{2n(n + 1)}{2} & x & n(n + 1) \\ \frac{6n(n + 1)(2n + 1)}{6} - n & y & n^{2}(2n + 3) \\ 4\left( \frac{n(n + 1)}{2} \right)^{2} - 2n\left( \frac{n(n + 1)}{2} \right) & y & n^{3}(n + 1) \end{matrix} \right|$$ = $\left| \begin{matrix} n(n + 1) & x & n(n + 1) \\ n^{2}(2n + 3) & y & n^{2}(2n + 3) \\ n^{3}(n + 1) & z & n^{3}(n + 1) \end{matrix} \right| = 0$\\ independent of x, y, z and n.