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

Answered by mahdipoor last updated on 07/Oct/23

=Σ_(x=−1) ^(−∞) (Σ_(y=x−1) ^(−∞) 2^(x+1) ×3^y ×5)=5(Σ_(x=−1) ^(−∞) 2^(x+1) (Σ_(y=x−1) ^(−∞) 3^y ))= 5(Σ_(x=−1) ^(−∞) 2^(x+1) (3^(x−1) +3^(x−2) +...+3^(−∞) ))= 5(Σ_(x=−1) ^(−∞) 2^(x+1) ×(((3^(x−1) +...+3^(−∞) )(3−1))/((3−1))))= 5(Σ_(x=−1) ^(−∞) 2^(x+1) ×(3^x /2))=5(Σ_(x=−1) ^(−∞) 6^x )=5×(6^0 /(6−1))=1

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

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