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Question Number 116056 by Study last updated on 30/Sep/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{5^n}{n!}=?$$

Answered by Dwaipayan Shikari last updated on 30/Sep/20

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}={\mathrm{e}}^{\mathrm{x}}-1$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{5^{\mathrm{n}}}{\mathrm{n}!}={\mathrm{e}}^5-1$$

Answered by MWSuSon last updated on 30/Sep/20

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{5^{\mathrm{n}}}{\mathrm{n}!}=(\frac{5}{1}+\frac{5^2}{2!}+\frac{5^3}{3!}+\frac{5^4}{4!}+...)$$$${\mathrm{e}}^{\mathrm{x}}=1+(\frac{\mathrm{x}}{1}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^4}{4!}+...)$$$$\mathrm{they}\mathrm{are}\mathrm{thesame}\mathrm{when}\mathrm{x}=5$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{5^{\mathrm{n}}}{\mathrm{n}!}={\mathrm{e}}^5-1$$$$$$

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