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Let-D-r-determinant-2-r-1-2-3-r-1-4-5-r-1-2-n-1-3-n-1-5-n-1-Then-the-value-of-r-1-n-D-r-is-




Question Number 15192 by arnabpapu550@gmail.com last updated on 08/Jun/17
Let D_r = determinant ((2^(r−1) ,(2 ∙ 3^(r−1) ),(4 ∙ 5^(r−1) )),((   α),(   β),(     γ)),((2^n −1),(3^n −1),(  5^n −1))).  Then the value of  Σ_(r=1) ^n D_r   is
$$\mathrm{Let}\:{D}_{{r}} =\begin{vmatrix}{\mathrm{2}^{{r}−\mathrm{1}} }&{\mathrm{2}\:\centerdot\:\mathrm{3}^{{r}−\mathrm{1}} }&{\mathrm{4}\:\centerdot\:\mathrm{5}^{{r}−\mathrm{1}} }\\{\:\:\:\alpha}&{\:\:\:\beta}&{\:\:\:\:\:\gamma}\\{\mathrm{2}^{{n}} −\mathrm{1}}&{\mathrm{3}^{{n}} −\mathrm{1}}&{\:\:\mathrm{5}^{{n}} −\mathrm{1}}\end{vmatrix}. \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{D}_{{r}} \:\:\mathrm{is} \\ $$

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