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Question Number 197776 by universe last updated on 28/Sep/23

Commented by witcher3 last updated on 28/Sep/23

$$\mid \underset{--}{\mathrm{n}}\mathrm{what}\mathrm{is}?$$

Commented by universe last updated on 28/Sep/23

$$\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{symbol}\mathrm{of}\mathrm{factorial}$$$$$$

Commented by mr W last updated on 28/Sep/23

$$somethingiswronginthequestion:$$$$\frac{2}{0!\times 2}+\frac{2^3}{2!\times 4}+\frac{2^5}{4!\times 6}+\frac{2^7}{6!\times 8}+...+\frac{2^{2n+1}}{\left(2n\right)!\times (2n+2)}+...$$

Answered by mr W last updated on 28/Sep/23

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

Commented by universe last updated on 28/Sep/23

$$thankssir$$

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