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Question Number 159775 by Ghaniy last updated on 22/Nov/21

$$\prod _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\alpha }^3+{\beta }^2}{3^{\mathrm{n}}}=?$$$$\mathrm{in}\mathrm{expanded}\mathrm{form}$$

Answered by Canebulok last updated on 21/Nov/21

$$$$$$\mathbf{Solution}:$$$$$$$$({\mathrm{x}}^3+{\pi }^2)\cdot \underset{\mathrm{n}=1}{\prod ^{100}}\frac{1}{4^{\mathrm{n}}}=\underset{\mathrm{n}=1}{\prod ^{100}}\frac{{\mathrm{x}}^3+{\pi }^2}{4^{\mathrm{n}}}$$$$$$$$\therefore$$$$({\mathrm{x}}^3+{\pi }^2)\cdot \underset{\mathrm{n}=1}{\prod ^{100}}\frac{1}{4^{\mathrm{n}}}=\frac{{\mathrm{x}}^3+{\pi }^2}{4^{\left(\underset{\mathrm{n}=1}{\sum ^{100}}\mathrm{n}\right)}}=\frac{{\mathrm{x}}^3+{\pi }^2}{4^{5050}}$$$$$$$$$$

Commented by Ghaniy last updated on 22/Nov/21

$$\mathrm{Thanks}\mathrm{sir}....\mathrm{I}{\mathrm{wasn}}^{\prime }\mathrm{t}\mathrm{sure}$$$$\mathrm{I}\mathrm{thought}\mathrm{the}100\mathrm{will}\mathrm{affect}({\alpha }^3+{\beta }^2)\mathrm{as}\mathrm{in}\frac{{({\alpha }^3+{\beta }^2)}^{\mathrm{\infty }}}{3^{5050}}$$$$$$

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