Given ${\mathrm{△}}_{\mathrm{r}}=\mid \begin{array}{ccc}{2}^{\mathrm{r}-1}& 2.3^{\mathrm{r}-1}& 4.5^{\mathrm{r}-1}\\ \mathrm{x}& \mathrm{y}& \mathrm{z}\\ 2^{\mathrm{n}-1}& 3^{\mathrm{n}-1}& 5^{\mathrm{n}-1}\end{array}\mid$ 

As $\sum _{\mathrm{r}=1}^{\mathrm{n}}\left({\mathrm{△}}_{\mathrm{r}}\right)$

$\Rightarrow \sum _{\mathrm{r}=1}^{\mathrm{n}} 2^{\mathrm{r}-1}=2^0+2^1+2^2+2^3+\cdots \cdot +2^{\mathrm{n}}$

$=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}$

$=\mathrm{i}\frac{\left(2^{\mathrm{n}}-1\right)}{2-1}$

= 2ⁿ – 1

$\Rightarrow \sum _{\mathrm{r}-1}^{\mathrm{n}} 2\cdot 3^{\mathrm{r}-1}=2\cdot \left[1+3+3^2+3^3+\cdots 3^{\mathrm{n}}\right]$

$=2\cdot \left(\frac{3^{\mathrm{n}}-1}{3-1}\right)$

$=2\times \frac{\left(3^{\mathrm{n}}-1\right)}{2}$

= 3ⁿ – 1

$\Rightarrow \sum _{\mathrm{r}-1}^{\mathrm{n}} 4.5^{\mathrm{r}-1}=4\cdot \left[5^0+5^1+5^2+5^3+\cdots \cdot 5^{\mathrm{n}}\right]$

$=4\cdot \frac{5^{\mathrm{n}}-1}{5-1}$

$=4\cdot \frac{5^{\mathrm{n}}-1}{4}$

$\therefore {\mathrm{D}}_{\mathrm{r}}=\left[\begin{array}{ccc}{2}^{\mathrm{n}}-1& 3^{\mathrm{n}}-1& 5^{\mathrm{n}}-1\\ \mathrm{x}& \mathrm{y}& \mathrm{z}\\ 2^{\mathrm{n}}-1& 3^{\mathrm{n}}-1& 5^{\mathrm{n}}-1\end{array}\right]$

∵ 2 column of determinant are equal

∴ D_r = xyz

∴ option A is correct