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Question Number 186689 by SANOGO last updated on 08/Feb/23

$$\underset{n=o}{\sum ^{+oo}}\frac{x^{3n}}{\left(2n\right)!}=?$$

Commented by mr W last updated on 08/Feb/23

$$whynot\mathrm{\infty }insteadofoo?$$$$whynot0insteadofo?$$

Answered by mr W last updated on 08/Feb/23

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+...\left(i\right)$$$$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+...\left(ii\right)$$$$\frac{\left(i\right)+\left(ii\right)}{2}:$$$$\frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+...=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{\left(2n\right)!}$$$$replacexwith{x}^{\frac{3}{2}}$$$$\frac{e^{\frac{3x}{2}}+e^{-\frac{3x}{2}}}{2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{3n}}{\left(2n\right)!}$$$$or$$$$\cosh \left(\frac{3x}{2}\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{3n}}{\left(2n\right)!}✓$$

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