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Question Number 167003 by cortano1 last updated on 04/Mar/22

$$\underset{\mathrm{n}=2}{\sum ^{\mathrm{n}=\mathrm{\infty }}}{\left(\frac{3}{4}\right)}^{\mathrm{n}}\cos \left(180°\mathrm{n}\right)=?$$

Answered by greogoury55 last updated on 04/Mar/22

$$\cos \left(180°n\right)={(-1)}^n$$$$\underset{n=2}{\sum ^{n=\mathrm{\infty }}}{\left(\frac{3}{4}\right)}^n{(-1)}^n=\underset{n=2}{\sum ^{\mathrm{\infty }}}{(-\frac{3}{4})}^n$$$$={(-\frac{3}{4})}^2+{(-\frac{3}{4})}^3+...$$$$=\frac{\frac{9}{16}}{1-(-\frac{3}{4})}=\frac{\frac{9}{16}}{\frac{7}{4}}=\frac{9}{16}\times \frac{4}{7}=\frac{9}{28}$$

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