Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com