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Question Number 130290 by bramlexs22 last updated on 24/Jan/21

$$\underset{\mathrm{i}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{j}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{i}+\mathrm{j}+\mathrm{k}}}?\mathrm{where}\mathrm{i}\ne \mathrm{j}\ne \mathrm{k}$$

Answered by EDWIN88 last updated on 24/Jan/21

$$\mathrm{S}=\underset{\mathrm{i}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{j}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{i}}.3^{\mathrm{j}}.3^{\mathrm{k}}}=\underset{\mathrm{i},\mathrm{j},\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{i}}{3}^{\mathrm{j}}{3}^{\mathrm{k}}}-3\underset{\mathrm{i},\mathrm{j},\mathrm{k}=\underset{\mathrm{i}=\mathrm{j}}{0}}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{i}}{3}^{\mathrm{j}}{3}^{\mathrm{k}}}+2\underset{\mathrm{i},\mathrm{j},\mathrm{k}=\underset{\mathrm{i}=\mathrm{j}=\mathrm{k}}{0}}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{i}}{3}^{\mathrm{j}}{3}^{\mathrm{k}}}$$$$\mathrm{S}={\left(\underset{\mathrm{i}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{i}}}\right)}^3-3\left(\underset{\mathrm{i}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{2\mathrm{i}}}\right)\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{\mathrm{k}}}\right)+2\underset{\mathrm{i}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{3\mathrm{i}}}$$$$\mathrm{S}={\left(\frac{1}{1-\frac{1}{3}}\right)}^3-3\left(\frac{1}{1-\frac{1}{3^2}}\right)\left(\frac{1}{1-\frac{1}{3}}\right)+2\left(\frac{1}{1-\frac{1}{3^3}}\right)$$$$\mathrm{S}=\frac{27}{8}-3.\frac{9}{8}.\frac{3}{2}+2.\frac{27}{26}=\frac{81}{208}$$

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