A) 1
B) - 1
C) 0
D) None of these
Correct Answer: C
Solution :
\[{{D}_{r}}=\left| \,\begin{matrix} {{2}^{r-1}} & {{2.3}^{r-1}} & {{4.5}^{r-1}} \\ x & y & z \\ {{2}^{n}}-1 & {{3}^{n}}-1 & {{5}^{n}}-1 \\ \end{matrix}\, \right|\] \[\Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}=\left| \,\begin{matrix} \sum\limits_{r=1}^{n}{{{2}^{r-1}}} & \sum\limits_{r=1}^{n}{{{2.3}^{r-1}}} & \sum\limits_{r=1}^{n}{{{4.5}^{r-1}}} \\ x & y & z \\ {{2}^{n}}-1 & {{3}^{n}}-1 & {{5}^{n}}-1 \\ \end{matrix} \right|}\] \[\Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}=}\left| \,\begin{matrix} {{2}^{n}}-1 & {{3}^{n}}-1 & {{5}^{n}}-1 \\ x & y & z \\ {{2}^{n}}-1 & {{3}^{n}}-1 & {{5}^{n}}-1 \\ \end{matrix} \right|\] Since we know that \[\sum\limits_{r=1}^{n}{{{2}^{r-1}}=\frac{{{2}^{n}}-1}{2-1}={{2}^{n}}-1,}\] \[2\sum\limits_{r=1}^{n}{{{3}^{r-1}}=2\frac{{{3}^{n}}-1}{3-1}={{3}^{n}}-1}\] and \[4\sum\limits_{r=1}^{n}{{{5}^{r-1}}=4\frac{{{5}^{n}}-1}{5-1}={{5}^{n}}-1}\] \[\Rightarrow \,\,\sum\limits_{r=1}^{n}{{{D}_{r}}=0}\], \[(\because {{R}_{1}}\equiv {{R}_{3}})\].
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