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Question Number 121919 by talminator2856791 last updated on 12/Nov/20

$$$$$$$$$$$$$$\underset{k=p}{\sum ^{\mathrm{\infty }}}4\cdot 3^{2-k}=\frac{2}{9}$$$$p=?$$$$$$$$$$

Answered by Dwaipayan Shikari last updated on 12/Nov/20

$$\underset{k=p}{\sum ^{\mathrm{\infty }}}4.3^{2-k}=4(3^{2-p}+3^{2-(p+1)}+...)$$$$=36(\frac{1}{3^p}+\frac{1}{3^{p+1}}+...)$$$$=36\left(\frac{\frac{1}{3^p}}{1-\frac{1}{3}}\right)=\frac{54}{3^p}=\frac{2}{9}\Rightarrow \frac{27}{3^{p-2}}=1$$$$3^{p-2}=3^3\Rightarrow p=5$$

Answered by geoplitz last updated on 12/Nov/20

$$\underset{k=p}{\sum ^{\mathrm{\infty }}}4\cdot 3^{2-k}=\frac{2}{9}$$$$4\underset{k=p}{\sum ^{\mathrm{\infty }}}{3}^2\cdot 3^{-k}=\frac{2}{9}$$$$4\cdot 3^2\underset{k=p}{\sum ^{\mathrm{\infty }}}{3}^{-k}=\frac{2}{9}$$$$36\underset{k=p}{\sum ^{\mathrm{\infty }}}{3}^{-k}=\frac{2}{9}$$$$\underset{k=p}{\sum ^{\mathrm{\infty }}}{3}^{-k}=\frac{2}{9\cdot 36}=\frac{2}{324}$$$$\frac{3^{-p+1}}{2}=\frac{2}{324}$$$$3^{-p+1}=\frac{4}{324}={\left(\frac{2}{18}\right)}^2=\frac{1}{81}$$$$3^{-p+1}=\frac{1}{81}=\frac{1}{3^4}=3^{-4}$$$$-p+1=-4$$$$-p=-5\Rightarrow p=5$$

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