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

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

prove that Σ_(x=0) ^∞ ((4^x . x)/(x!)) = 4 e^4

$${prove}\:{that}\:\underset{{x}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{4}^{{x}} \:.\:{x}}{{x}!}\:=\:\mathrm{4}\:{e}^{\mathrm{4}} \\ $$

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

Σ_(x=0) ^∞ ((4^x x)/(x!))=Σ_(x=1) ^∞ ((4^x x)/(x!))=4Σ_(x=1) ^∞ (4^(x−1) /((x−1)!))=4Σ_(x=0) ^∞ (4^x /(x!))=4e^4

$$\underset{{x}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{{x}} {x}}{{x}!}=\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{{x}} {x}}{{x}!}=\mathrm{4}\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{{x}−\mathrm{1}} }{\left({x}−\mathrm{1}\right)!}=\mathrm{4}\underset{{x}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{{x}} }{{x}!}=\mathrm{4}{e}^{\mathrm{4}} \\ $$

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

Σ_(n=0) ^∞ (x^n /(n!))=e^x Σ_(n=0) ^∞ ((nx^(n−1) )/(n!))=e^x ⇒Σ_(n=0) ^∞ ((nx^n )/(n!))=xe^x set x=4: Σ_(n=0) ^∞ ((4^n n)/(n!))=4e^4 or Σ_(x=0) ^∞ ((4^x x)/(x!))=4e^4

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

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