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# If $D_r=\begin{vmatrix}2^{r-1}&2.3^{r-1}&4.5^{r-1}\\\alpha&\beta&\gamma\\2^n-1&3^n-1&5^n-1\end{vmatrix}$ then the value of $\sum\limits_{r=1}^n D_r$ is

$\begin{array}{1 1}(A)\;0&(B)\;\alpha\beta\gamma\\(C)\;\alpha+\beta+\gamma&(D)\;\alpha.2^n+\beta.3^n+\gamma.4^n\end{array}$

Can you answer this question?

$\sum\limits_{r=1}^n2^{r-1}=1+2+2^2+......+2^{n-1}$
$\Rightarrow \large\frac{2^n-1}{2-1}$$=2^n-1$
$\sum\limits_{r=1}^n2.3^{r-1}=\large\frac{2(3^n-1)}{3-1}$
$\Rightarrow 3^n-1$
$\sum\limits_{r=1}^n4.5^{n-1}=\large\frac{4(5^n-1)}{5-1}$
$\Rightarrow 5^n-1$
$\therefore \sum\limits_{r=1}^n D_r=\begin{vmatrix}2^n-1&3^n-1&5^n-1\\\alpha&\beta&\gamma\\2^n-1&3^n-1&5^n-1\end{vmatrix}=0$
Hence (A) is the correct answer.
answered Apr 24, 2014