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Question Number 198014 by hardmath last updated on 07/Oct/23

Answered by mahdipoor last updated on 07/Oct/23

$$=\underset{x=-1}{\sum ^{-\mathrm{\infty }}}\left(\underset{y=x-1}{\sum ^{-\mathrm{\infty }}}{2}^{x+1}\times 3^y\times 5\right)=5\left(\underset{x=-1}{\sum ^{-\mathrm{\infty }}}{2}^{x+1}\left(\underset{y=x-1}{\sum ^{-\mathrm{\infty }}}{3}^y\right)\right)=$$$$5\left(\underset{x=-1}{\sum ^{-\mathrm{\infty }}}{2}^{x+1}(3^{x-1}+3^{x-2}+...+3^{-\mathrm{\infty }})\right)=$$$$5\left(\underset{x=-1}{\sum ^{-\mathrm{\infty }}}{2}^{x+1}\times \frac{(3^{x-1}+...+3^{-\mathrm{\infty }})(3-1)}{(3-1)}\right)=$$$$5\left(\underset{x=-1}{\sum ^{-\mathrm{\infty }}}{2}^{x+1}\times \frac{3^x}{2}\right)=5\left(\underset{x=-1}{\sum ^{-\mathrm{\infty }}}{6}^x\right)=5\times \frac{6^0}{6-1}=1$$

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