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Question Number 181360 by lapache last updated on 24/Nov/22

$$Calcul$$$$\underset{k=0}{\sum ^n}k{\left(\frac{1}{3}\right)}^k=...$$

Answered by mahdipoor last updated on 24/Nov/22

$$way2:$$$$getf\left(x\right)=e^x+e^{2x}+e^{3x}+...+e^{kx}$$$$f\left(x\right)(1-e^x)=e^x-e^{(k+1)x}\Rightarrow f\left(x\right)=\frac{e^x-e^{(k+1)x}}{1-e^x}$$$$\Rightarrow \Rightarrow$$$$\frac{df}{dx}=e^x+2e^{2x}+3e^{3x}+...+{ke}^{kx}$$$$\frac{df}{dx}=\frac{(e^x-(k+1)e^{(k+1)x})(1-e^x)-(-e^x)(e^x-e^{(k+1)x})}{{(1-e^x)}^2}=$$$$\Rightarrow x=ln3or{e}^x=\frac{1}{3}\Rightarrow$$$$\frac{df}{dx}(x=ln3)=\frac{1}{3}+\frac{2}{3^2}+....+\frac{k}{3^k}$$$$\frac{df}{dx}(x=ln3)=\frac{(\frac{1}{3}-\frac{k+1}{3^{k+1}})\left(\frac{2}{3}\right)-\left(\frac{-1}{3}\right)(\frac{1}{3}-\frac{1}{3^{k+1}})}{{\left(\frac{2}{3}\right)}^2}=$$$$\frac{3}{4}-\frac{2k+3}{4\times 3^k}$$$$\Rightarrow \Rightarrow \frac{1}{3}+\frac{2}{3^2}+...+\frac{k}{3^k}=\frac{3}{4}-\frac{2k+3}{4\times 3^k}$$

Answered by SEKRET last updated on 24/Nov/22

$$\mathbf{A}=\underset{\mathrm{k}=0}{\sum ^{\mathbf{n}}}\frac{\mathbf{k}}{3^{\mathbf{k}}}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+....+\frac{\mathbf{n}}{3^{\mathbf{n}}}$$$$\frac{\mathbf{A}}{3}=\underset{\mathbf{k}=0}{\sum ^{\mathbf{n}}}\frac{\mathbf{k}}{3^{\mathbf{k}+1}}=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+..+\frac{\mathbf{n}}{3^{\mathbf{n}+1}}$$$$\frac{2\mathbf{A}}{3}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{\mathbf{n}}}+\frac{\mathbf{n}}{3^{\mathbf{n}+1}}$$$$......$$$$$$

Answered by mahdipoor last updated on 24/Nov/22

$$I\Rightarrow (\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...\frac{n}{3^n})\times (1-\frac{1}{3})=$$$$\frac{1}{3}+\frac{2}{3^2}+...+\frac{n-1}{3^{n-1}}+\frac{n}{3^n}-(\frac{1}{3^2}+\frac{2}{3^3}+...\frac{n-1}{3^n}+\frac{n}{3^{n+1}})=$$$$\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n}-\frac{n}{3^{n+1}}$$$$II\Rightarrow (\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n})\times (1-\frac{1}{3})=\frac{1}{3}-\frac{1}{3^{n+1}}$$$$I\mathrm{\&}II\Rightarrow \underset{k=0}{\sum ^n}k{\left(\frac{1}{3}\right)}^k=(\frac{1}{3}+\frac{2}{3^2}+...+\frac{n}{3^n})=$$$$\frac{(\frac{1}{3}+\frac{2}{3^2}+...+\frac{n}{3^n})(1-\frac{1}{3})}{(1-\frac{1}{3})}=\frac{\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n}-\frac{n}{3^{n+1}}}{\frac{2}{3}}=$$$$\frac{\frac{(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n})(1-\frac{1}{3})}{(1-\frac{1}{3})}-\frac{n}{3^{n+1}}}{\frac{2}{3}}=$$$$\frac{\frac{(\frac{1}{3}-\frac{1}{3^{n+1}})}{\frac{2}{3}}-\frac{n}{3^{n+1}}}{\frac{2}{3}}=\frac{3}{4}-\frac{3+2n}{4\times 3^n}$$

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