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Question Number 86598 by Rio Michael last updated on 29/Mar/20

$$\mathrm{write}\mathrm{out}\mathrm{the}\mathrm{general}\mathrm{summation}\mathrm{formula}\mathrm{for}$$$$\mathrm{the}\mathrm{maclaurin}\mathrm{series}\mathrm{expansion}\mathrm{for}\frac{1}{2}(\cos x+\cosh x)$$

Commented by mathmax by abdo last updated on 31/Mar/20

$$cosx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)!}{x}^{2n}andch\left(x\right)=\frac{e^x+e^{-x}}{2}$$$$=\frac{1}{2}(\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}+\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}{x}^n)$$$$=\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\{1+{(-1)}^n\}{x}^n=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n}\Rightarrow }{\left(2n\right)!}$$$$\frac{1}{2}(cosx+ch\left(x\right))=\frac{1}{2}\{\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)!}{x}^{2n}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\left(2n\right)!}{x}^{2n}\}$$$$=\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{1+{(-1)}^n}{\left(2n\right)!}{x}^{2n}=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{4n}}{\left(4n\right)!}$$

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