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Question Number 145304 by imjagoll last updated on 04/Jul/21

$$1+\frac{3\mathrm{x}}{1!}+\frac{{5\mathrm{x}}^2}{2!}+\frac{{7\mathrm{x}}^3}{3!}+\frac{{9\mathrm{x}}^4}{4!}+...+\mathrm{\infty }=?$$

Answered by qaz last updated on 04/Jul/21

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{(2\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}=(2\mathrm{xD}+1)\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}=(2\mathrm{xD}+1){\mathrm{e}}^{\mathrm{x}}=(2\mathrm{x}+1){\mathrm{e}}^{\mathrm{x}}$$

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