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Question Number 165262 by mkam last updated on 28/Jan/22

$$\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{u}\mathit{m}\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{2}\right)}^{\mathit{n}}+\mathit{i}{\left(\frac{1}{3}\right)}^{\mathit{n}}$$

Answered by Mathspace last updated on 28/Jan/22

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n+i{\left(\frac{1}{3}\right)}^n\}=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n-1$$$$+i\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{3}\right)}^n-i$$$$=\frac{1}{1-\frac{1}{2}}-1+i\frac{1}{1-\frac{1}{3}}-i$$$$=1+i\frac{3}{2}-i=1+i(\frac{3}{2}-1)$$$$=1+\frac{1}{2}i$$

Answered by Ar Brandon last updated on 28/Jan/22

$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}[{\left(\frac{1}{2}\right)}^n+i{\left(\frac{1}{3}\right)}^n]$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{2}\right)}^n+i\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{3}\right)}^n$$$$=1+i\frac{1}{2}$$

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