prove-that-x-0-4-x-x-x-4-e-4-

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

p r o v e t h a t \underset{x = 0}{\sum^{\infty}} \frac{4^{x} . x}{x!} = 4 e^{4}

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

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

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

\underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} = e^{x}

\underset{n = 0}{\sum^{\infty}} \frac{{n x}^{n - 1}}{n!} = e^{x}

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{{n x}^{n}}{n!} = {x e}^{x}

s e t x = 4 :

\underset{n = 0}{\sum^{\infty}} \frac{4^{n} n}{n!} = 4 e^{4}

o r

\underset{x = 0}{\sum^{\infty}} \frac{4^{x} x}{x!} = 4 e^{4}

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