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Question Number 200446 by hardmath last updated on 18/Nov/23

$$fund\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(3+{(-1)}^{\mathit{n}})}^{\mathit{n}}}{n}{x}^{\mathit{n}}=?$$

Answered by mr W last updated on 19/Nov/23

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{2n}=\frac{x^2}{1-x^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^x{x}^{2n}dx={\int }_0^x\frac{x^{2}dx}{1-x^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{2n+1}}{2n+1}dx={\int }_0^x(\frac{1}{1-x^2}-1)dx=\frac{1}{2}\ln \frac{1+x}{1-x}-x$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n+1}}{2n+1}dx=\frac{1}{2}\ln \frac{1+x}{1-x}$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(2x\right)}^{2n+1}}{2n+1}dx=\frac{1}{2}\ln \frac{1+2x}{1-2x}$$$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{2n-1}=\frac{x}{1-x^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^3{x}^{2n-1}dx={\int }_0^x\frac{x}{1-x^2}dx$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{2n}=\frac{1}{2}\ln \frac{1}{1-x^2}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\left(4x\right)}^{2n}}{2n}=\frac{1}{2}\ln \frac{1}{1-16x^2}$$$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(3+{(-1)}^n)}^n{x}^n}{n}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2^{2n+1}{x}^{2n+1}}{2n+1}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{4^{2n}{x}^{2n}}{2n}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(2x\right)}^{2n+1}}{2n+1}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\left(4x\right)}^{2n}}{2n}$$$$=\frac{1}{2}\ln \frac{1+2x}{1-2x}+\frac{1}{2}\ln \frac{1}{1-16x^2}$$$$=\frac{1}{2}\ln \frac{1+2x}{(1-2x)(1-16x^2)}✓$$

Commented by hardmath last updated on 19/Nov/23

$$perfectmydearprofessorthankyousomuch$$

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