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Question Number 162190 by naka3546 last updated on 27/Dec/21

$$\lfloor \frac{3^2+1}{3^2-1}+\frac{3^3+1}{3^3-1}+\frac{3^4+1}{3^4-1}+\dots +\frac{3^{2017}+1}{3^{2017}-1}\rfloor =?$$

Answered by mr W last updated on 28/Dec/21

$$\underset{k=2}{\sum ^n}\frac{3^k+1}{3^k-1}$$$$=\underset{k=2}{\sum ^n}(1+\frac{2}{3^k-1})$$$$=n-1+\underset{k=1}{\sum ^n}\frac{2}{3^k-1}-1$$$$=n-2+\underset{k=1}{\sum ^n}\frac{2}{3^k-1}$$$$=n-2+S_{n}with{S}_n=\underset{k=1}{\sum ^n}\frac{2}{3^k-1}$$$$\lfloor \underset{k=2}{\sum ^n}\frac{3^k+1}{3^k-1}\rfloor =n-2+\lfloor S_n\rfloor$$$$S_1⩽S_n<S_{\mathrm{\infty }}$$$$S_1=1$$$$S_{\mathrm{\infty }}=\underset{n\to 0}{\lim }\underset{k=1}{\sum ^n}\frac{2}{3^k-1}=2\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{3^k-1}=2\times \frac{\ln 3-\ln 2-{\psi }_{1/3}\left(1\right)}{\ln 3}\approx 1.36431$$$$1⩽S_n<1.36431$$$$\Rightarrow \lfloor S_n\rfloor =1$$$$\lfloor \underset{k=2}{\sum ^n}\frac{3^k+1}{3^k-1}\rfloor =n-2+\lfloor S_n\rfloor =n-2+1=n-1$$$$\Rightarrow \lfloor \underset{k=2}{\sum ^n}\frac{3^k+1}{3^k-1}\rfloor =n-1$$$$forn=2017:$$$$\Rightarrow \lfloor \underset{k=2}{\sum ^{2017}}\frac{3^k+1}{3^k-1}\rfloor =2017-1=2016$$

Commented by naka3546 last updated on 28/Dec/21

$$Thankyou,sir.$$

Commented by mr W last updated on 28/Dec/21

$$\underset{\mathit{n}=1}{\mathbf{\sum }^{\mathrm{\infty }}}\frac{1}{{\mathit{k}}^{\mathit{n}}-1}=\frac{\mathbf{ln}\mathit{k}-\mathbf{ln}(\mathit{k}-1)-{\mathit{\psi }}_{1/\mathit{k}}\left(1\right)}{\mathbf{ln}\mathit{k}}$$

Commented by Rasheed.Sindhi last updated on 28/Dec/21

$${\mathit{e}}^{\mathit{x}}cellentsir!$$

Commented by Tawa11 last updated on 28/Dec/21

$$\mathrm{Great}\mathrm{sir}.$$

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