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Question Number 209800 by depressiveshrek last updated on 22/Jul/24

Commented by depressiveshrek last updated on 22/Jul/24

$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{this}\mathrm{series}$$

Answered by mr W last updated on 22/Jul/24

$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{3^2}{\left(\frac{3\sqrt{2}}{{xy}^2\sqrt[3]{z}}\right)}^n(n-1)$$$$=\frac{1}{9}\underset{n=1}{\sum ^{\mathrm{\infty }}}(n-1)k^{n}withk=\frac{3\sqrt{2}}{{xy}^2\sqrt[3]{z}}$$$$assume\mid k\mid <1,otherwise{it}^{\prime }snot$$$$convergent.$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{n-1}=\frac{1}{1-x}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}(n-1)x^{n-2}=\frac{1}{{(1-x)}^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}(n-1)x^n=\frac{x^2}{{(1-x)}^2}={\left(\frac{1}{1-\frac{1}{x}}\right)}^2$$$$setx=k:$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}(n-1)k^n={\left(\frac{1}{1-\frac{1}{k}}\right)}^2={\left(\frac{1}{1-\frac{{xy}^2\sqrt[3]{z}}{3\sqrt{2}}}\right)}^2$$$$\frac{1}{9}\underset{n=1}{\sum ^{\mathrm{\infty }}}(n-1)k^n={\left(\frac{1}{3-\frac{{xy}^2\sqrt[3]{z}}{\sqrt{2}}}\right)}^2$$$$=\frac{2}{{(3\sqrt{2}-{xy}^2\sqrt[3]{z})}^2}✓$$

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