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Question-197776

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Question Number 197776 by universe last updated on 28/Sep/23
Commented by witcher3 last updated on 28/Sep/23
∣n_(−−) what is ?
$$\mid\underset{−−} {\mathrm{n}}\:\:\:\mathrm{what}\:\mathrm{is}\:? \\ $$
Commented by universe last updated on 28/Sep/23
this is a symbol of factorial
$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{symbol}\:\mathrm{of}\:\mathrm{factorial} \\ $$$$ \\ $$
Commented by mr W last updated on 28/Sep/23
something is wrong in the question: (2/(0!×2))+(2^3 /(2!×4))+(2^5 /(4!×6))+(2^7 /(6!×8))+...+(2^(2n+1) /((2n)!×(2n+2)))+...
$${something}\:{is}\:{wrong}\:{in}\:{the}\:{question}: \\ $$$$\frac{\mathrm{2}}{\mathrm{0}!×\mathrm{2}}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!×\mathrm{4}}+\frac{\mathrm{2}^{\mathrm{5}} }{\mathrm{4}!×\mathrm{6}}+\frac{\mathrm{2}^{\mathrm{7}} }{\mathrm{6}!×\mathrm{8}}+…+\frac{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}\right)!×\left(\mathrm{2}{n}+\mathrm{2}\right)}+… \\ $$
Answered by mr W last updated on 28/Sep/23
e^x =Σ_(n=0) ^∞ (x^(2n) /((2n)!))+Σ_(n=0) ^∞ (x^(2n+1) /((2n+1)!)) e^(−x) =Σ_(n=0) ^∞ (x^(2n) /((2n)!))−Σ_(n=0) ^∞ (x^(2n+1) /((2n+1)!)) ((e^x +e^(−x) )/2)=Σ_(n=0) ^∞ (x^(2n) /((2n)!)) ((x(e^x +e^(−x) ))/2)=Σ_(n=0) ^∞ (x^(2n+1) /((2n)!)) ∫_0 ^x ((x(e^x +e^(−x) ))/2)dx=Σ_(n=0) ^∞ ∫_0 ^x (x^(2n+1) /((2n)!))dx (((x−1)e^x −(x+1)e^(−x) +2)/2)=Σ_(n=0) ^∞ (x^(2(n+1)) /(2(n+1)(2n)!)) (((x−1)e^x −(x+1)e^(−x) +2)/(2x))=Σ_(n=0) ^∞ (x^(2n+1) /(2(n+1)(2n)!)) let x=2 (((2−1)e^2 −(2+1)e^(−2) +2)/(2×2))=Σ_(n=0) ^∞ (2^(2n+1) /(2(n+1)(2n)!)) ⇒(1/2)+(e^2 /4)−((3e^(−2) )/4)=Σ_(n=0) ^∞ (2^(2n+1) /(2(n+1)(2n)!))
$${e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${e}^{−{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$\frac{{x}\left({e}^{{x}} +{e}^{−{x}} \right)}{\mathrm{2}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$\int_{\mathrm{0}} ^{{x}} \frac{{x}\left({e}^{{x}} +{e}^{−{x}} \right)}{\mathrm{2}}{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{{x}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}\right)!}{dx} \\ $$$$\frac{\left({x}−\mathrm{1}\right){e}^{{x}} −\left({x}+\mathrm{1}\right){e}^{−{x}} +\mathrm{2}}{\mathrm{2}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}\left({n}+\mathrm{1}\right)} }{\mathrm{2}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!} \\ $$$$\frac{\left({x}−\mathrm{1}\right){e}^{{x}} −\left({x}+\mathrm{1}\right){e}^{−{x}} +\mathrm{2}}{\mathrm{2}{x}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!} \\ $$$${let}\:{x}=\mathrm{2} \\ $$$$\frac{\left(\mathrm{2}−\mathrm{1}\right){e}^{\mathrm{2}} −\left(\mathrm{2}+\mathrm{1}\right){e}^{−\mathrm{2}} +\mathrm{2}}{\mathrm{2}×\mathrm{2}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}+\frac{{e}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{3}{e}^{−\mathrm{2}} }{\mathrm{4}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!} \\ $$
Commented by universe last updated on 28/Sep/23
thanks sir
$${thanks}\:{sir}\: \\ $$

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