Question

Let $D_r=\left|\begin{array}{ccc}{2}^{r-1}& {2.3}^{r-1}& {4.5}^{r-1}\\ \alpha & \beta & \gamma \\ 2^n-1& 3^n-1& 5^n-1\end{array}\right|$. Then, the value of ${\sum }_{r=1}^n{D}_r$ is

• A. $\alpha \beta \gamma$
• B. $2^n\alpha +3^n\beta +4^n\gamma$
• C. $2\alpha +3\beta +4\gamma$
• D. None of these

Answer 1

The correct option is D None of these

$D_r=\left[\begin{array}{ccc}{2}^{r-1}& {2-3}^{r-1}& {4-5}^{r-1}\\ \alpha & \beta & \gamma \\ 2^n-1& 3^n-1& 5^n-1\end{array}\right]$

${\sum }_{r=1}^n{D}_r$

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

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

$=1\frac{(2^n-1)}{2-1}$

$=2^n-1$

$\Rightarrow {\sum }_{r=1}^n{2.3}^{r-1}=2.[1+3+3^2+3^3+....{.3}^n]$

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

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

$=3^n-1$

$\Rightarrow {\sum }_{r=1}^n{4.5}^{r-1}=4.[5^0+5^1+5^2+5^3+...{.5}^n]$

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

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

$=5^n-1$

$\therefore D_r=\left[\begin{array}{ccc}{2}^n-1& 3^n-1& 5^n-1\\ \alpha & \beta & \gamma \\ 2^n-1& 3^n-1& 5^n-1\end{array}\right]=0$

$\because 2$ columns of determinant are equal.

$\therefore D_r=0$

Hence, solved.