Question Number 152435 by fotosy2k last updated on 28/Aug/21 | ||
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Commented by fotosy2k last updated on 28/Aug/21 | ||
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$${pls}\:{help} \\ $$ | ||
Answered by qaz last updated on 28/Aug/21 | ||
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$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{n}} }\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\frac{\mathrm{1}}{\mathrm{5}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ | ||
Commented by fotosy2k last updated on 30/Aug/21 | ||
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$${thank}\:{you} \\ $$ | ||