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Question-165321

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Question Number 165321 by amin96 last updated on 29/Jan/22
Answered by mahdipoor last updated on 31/Jan/22
{ ((u^2 +2u+4⇒^(u=x) =x^2 +2x+4)),((u^2 −2u+4⇒^(u=x+2) x^2 +2x+1)) :}⇒ Π_(i=u) ^m (u^2 +2u+4)=Π_(i+2=u) ^(m+2) (u^2 −2u+4) { ((u−2⇒^(u=x+4) x+2)),((u+2⇒^(u=x) x+2)) :}⇒Π_(i+4=u) ^(m+4) u−2=Π_(i=u) ^m u+2 ⇒Π_(u=3) ^∞ ((u^3 −8)/(u^3 +8))=Π_(u=3) ^∞ (((u−2)(u^2 +2u+4))/((u+2)(u^2 −2u+4)))= ((1×2×3×4)/(7×12))×((Π_(u=7) ^∞ (u−2)×Π_3 ^∞ (u^2 +2u+4))/(Π_(u=3) ^∞ (u+2)×Π_5 ^∞ (u^2 −2u+4)))= (2/7)
$$\begin{cases}{{u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{4}\overset{{u}={x}} {\Rightarrow}={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}\\{{u}^{\mathrm{2}} −\mathrm{2}{u}+\mathrm{4}\overset{{u}={x}+\mathrm{2}} {\Rightarrow}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}\end{cases}\Rightarrow \\ $$$$\underset{{i}={u}} {\overset{{m}} {\prod}}\left({u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{4}\right)=\underset{{i}+\mathrm{2}={u}} {\overset{{m}+\mathrm{2}} {\prod}}\left({u}^{\mathrm{2}} −\mathrm{2}{u}+\mathrm{4}\right) \\ $$$$\begin{cases}{{u}−\mathrm{2}\overset{{u}={x}+\mathrm{4}} {\Rightarrow}{x}+\mathrm{2}}\\{{u}+\mathrm{2}\overset{{u}={x}} {\Rightarrow}{x}+\mathrm{2}}\end{cases}\Rightarrow\underset{{i}+\mathrm{4}={u}} {\overset{{m}+\mathrm{4}} {\prod}}{u}−\mathrm{2}=\underset{{i}={u}} {\overset{{m}} {\prod}}{u}+\mathrm{2} \\ $$$$\Rightarrow\underset{{u}=\mathrm{3}} {\overset{\infty} {\prod}}\frac{{u}^{\mathrm{3}} −\mathrm{8}}{{u}^{\mathrm{3}} +\mathrm{8}}=\underset{{u}=\mathrm{3}} {\overset{\infty} {\prod}}\frac{\left({u}−\mathrm{2}\right)\left({u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{4}\right)}{\left({u}+\mathrm{2}\right)\left({u}^{\mathrm{2}} −\mathrm{2}{u}+\mathrm{4}\right)}= \\ $$$$\frac{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}}{\mathrm{7}×\mathrm{12}}×\frac{\underset{{u}=\mathrm{7}} {\overset{\infty} {\prod}}\left({u}−\mathrm{2}\right)×\underset{\mathrm{3}} {\overset{\infty} {\prod}}\left({u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{4}\right)}{\underset{{u}=\mathrm{3}} {\overset{\infty} {\prod}}\left({u}+\mathrm{2}\right)×\underset{\mathrm{5}} {\overset{\infty} {\prod}}\left({u}^{\mathrm{2}} −\mathrm{2}{u}+\mathrm{4}\right)}= \\ $$$$\frac{\mathrm{2}}{\mathrm{7}} \\ $$
Commented by amin96 last updated on 30/Jan/22
(2/7)
$$\frac{\mathrm{2}}{\mathrm{7}} \\ $$
Answered by puissant last updated on 31/Jan/22
k=n+2 Π_(n=1) ^∞ (((n+2)^3 −2^3 )/((n+2)^2 +2^3 )) = Π_(k=3) ^∞ ((k^3 −2^3 )/(k^3 +2^3 )) = Π_(k=3) ^∞ (((k−2)(k^2 +2k+4))/((k+2)(k^2 −2k+4))) = lim_(k→∞) Π_(t=3) ^k (((t−2)/(t+2)))×lim_(k→∞) Π_(t=3) ^k ( ((t^2 +2t+4)/(t^2 −2t+4))) =lim_(k→∞) (1/5)•(2/6)•(3/7)•(4/8)•(5/9)•....•((k−2)/(k+2)) × lim_(k→∞) ((19)/7)•((28)/(12))•((39)/(19))•((52)/(28))•...•((k^2 +3)/(k^2 −4k+7))•((k^2 +2k+4)/(k^2 −2k+4)) =24 lim_(k→∞) (1/((k−1)k(k+1)(k+2))) × (1/(7×12))lim_(k→∞) (k^2 +3)(k^2 +2k+4) = ((24)/(84)) lim_(k→∞) (((k^2 +3)(k^2 +2k+4))/((k−1)k(k+1)(k+2))) = ((24)/(84)). Π_(n=1) ^∞ (((n+2)^3 −8)/((n+2)^3 +8)) = (2/7)... ...............Le puissant..............
$${k}={n}+\mathrm{2}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\left({n}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} }{\left({n}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} }\:=\:\underset{{k}=\mathrm{3}} {\overset{\infty} {\prod}}\frac{{k}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} }{{k}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} } \\ $$$$=\:\underset{{k}=\mathrm{3}} {\overset{\infty} {\prod}}\frac{\left({k}−\mathrm{2}\right)\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{4}\right)}{\left({k}+\mathrm{2}\right)\left({k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{4}\right)} \\ $$$$=\:\underset{{k}\rightarrow\infty} {\mathrm{lim}}\underset{{t}=\mathrm{3}} {\overset{{k}} {\prod}}\:\left(\frac{{t}−\mathrm{2}}{{t}+\mathrm{2}}\right)×\underset{{k}\rightarrow\infty} {\mathrm{lim}}\underset{{t}=\mathrm{3}} {\overset{{k}} {\prod}}\left(\:\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{4}}{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{4}}\right) \\ $$$$=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{5}}\bullet\frac{\mathrm{2}}{\mathrm{6}}\bullet\frac{\mathrm{3}}{\mathrm{7}}\bullet\frac{\mathrm{4}}{\mathrm{8}}\bullet\frac{\mathrm{5}}{\mathrm{9}}\bullet….\bullet\frac{{k}−\mathrm{2}}{{k}+\mathrm{2}}\:× \\ $$$$\underset{{k}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{19}}{\mathrm{7}}\bullet\frac{\mathrm{28}}{\mathrm{12}}\bullet\frac{\mathrm{39}}{\mathrm{19}}\bullet\frac{\mathrm{52}}{\mathrm{28}}\bullet…\bullet\frac{{k}^{\mathrm{2}} +\mathrm{3}}{{k}^{\mathrm{2}} −\mathrm{4}{k}+\mathrm{7}}\bullet\frac{{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{4}}{{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{4}} \\ $$$$=\mathrm{24}\:\underset{{k}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right){k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\:× \\ $$$$\frac{\mathrm{1}}{\mathrm{7}×\mathrm{12}}\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left({k}^{\mathrm{2}} +\mathrm{3}\right)\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{4}\right) \\ $$$$=\:\frac{\mathrm{24}}{\mathrm{84}}\:\underset{{k}\rightarrow\infty} {\mathrm{lim}}\frac{\left({k}^{\mathrm{2}} +\mathrm{3}\right)\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{4}\right)}{\left({k}−\mathrm{1}\right){k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\:=\:\frac{\mathrm{24}}{\mathrm{84}}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\left({n}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{8}}{\left({n}+\mathrm{2}\right)^{\mathrm{3}} +\mathrm{8}}\:=\:\frac{\mathrm{2}}{\mathrm{7}}… \\ $$$$\:\:\:\:\:\:\:\:\:\:……………\mathscr{L}{e}\:{puissant}………….. \\ $$

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