A) 0
B) 1
C) \[\frac{n(n+1)(2n+1)}{6}\]
D) none of these
Correct Answer: A
Solution :
\[\because \] \[{{D}_{r}}=\left| \begin{matrix} r & 1 & \frac{n(n+1)}{2} \\ 2r-1 & 4 & {{n}^{2}} \\ {{2}^{r-1}} & 5 & {{2}^{n}}-1 \\ \end{matrix} \right|\] \[\sum\limits_{r=0}^{n}{{{D}_{r}}}=\left| \begin{matrix} \sum{r} & 1 & \frac{n(n+1)}{2} \\ 2\sum{r}-\sum{1} & 4 & {{n}^{2}} \\ \sum{{{2}^{r-1}}} & 5 & {{2}^{n}}-1 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \frac{n(n+1)}{2} & 1 & \frac{n(n+1)}{2} \\ {{n}^{2}} & 4 & {{n}^{2}} \\ {{2}^{n}}-1 & 5 & {{2}^{n}}-1 \\ \end{matrix} \right|\] \[=0\](\[\because \]two columns are identical
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