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Question Number 183420 by mokys last updated on 25/Dec/22

$$provethat\underset{x=0}{\sum ^{\mathrm{\infty }}}\frac{4^x.x}{x!}=4e^4$$

Answered by Ar Brandon last updated on 25/Dec/22

$$\underset{x=0}{\sum ^{\mathrm{\infty }}}\frac{4^{x}x}{x!}=\underset{x=1}{\sum ^{\mathrm{\infty }}}\frac{4^{x}x}{x!}=4\underset{x=1}{\sum ^{\mathrm{\infty }}}\frac{4^{x-1}}{(x-1)!}=4\underset{x=0}{\sum ^{\mathrm{\infty }}}\frac{4^x}{x!}=4e^4$$

Answered by mr W last updated on 26/Dec/22

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}=e^x$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{nx}^{n-1}}{n!}=e^x$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{nx}^n}{n!}={xe}^x$$$$setx=4:$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{4^{n}n}{n!}=4e^4$$$$or$$$$\underset{x=0}{\sum ^{\mathrm{\infty }}}\frac{4^{x}x}{x!}=4e^4$$

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