(c) ${D_r} = \left| {\,\begin{array}{*{20}{c}}{{2^{r – 1}}}&{{{2.3}^{r – 1}}}&{{{4.5}^{r – 1}}}\\x&y&z\\{{2^n} – 1}&{{3^n} – 1}&{{5^n} – 1}\end{array}\,} \right|$

$ \Rightarrow \sum\limits_{r = 1}^n {{D_r} = \left| {\,\begin{array}{*{20}{c}}{\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{array}} \right|} $

$ \Rightarrow \sum\limits_{r = 1}^n {{D_r} = } \left| {\,\begin{array}{*{20}{c}}{{2^n} – 1}&{{3^n} – 1}&{{5^n} – 1}\\x&y&z\\{{2^n} – 1}&{{3^n} – 1}&{{5^n} – 1}\end{array}} \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} $, $(\because {R_1} \equiv {R_3})$ .