Math Question and Answers


Question and Answers Forum

All Questions Topic List
Set Theory Questions
Previous in All Question Next in All Question
Previous in Set Theory Next in Set Theory
Question Number 165321 by amin96 last updated on 29/Jan/22

	Answered by mahdipoor last updated on 31/Jan/22
	${ ((u^2 +2u+4⇒^(u=x) =x^2 +2x+4)),((u^2 −2u+4⇒^(u=x+2) x^2 +2x+1)) :}⇒ Π_(i=u) ^m (u^2 +2u+4)=Π_(i+2=u) ^(m+2) (u^2 −2u+4) { ((u−2⇒^(u=x+4) x+2)),((u+2⇒^(u=x) x+2)) :}⇒Π_(i+4=u) ^(m+4) u−2=Π_(i=u) ^m u+2 ⇒Π_(u=3) ^∞ ((u^3 −8)/(u^3 +8))=Π_(u=3) ^∞ (((u−2)(u^2 +2u+4))/((u+2)(u^2 −2u+4)))= ((1×2×3×4)/(7×12))×((Π_(u=7) ^∞ (u−2)×Π_3 ^∞ (u^2 +2u+4))/(Π_(u=3) ^∞ (u+2)×Π_5 ^∞ (u^2 −2u+4)))= (2/7)$
	${\begin{cases} u^{2} + 2 u + 4 \overset{u = x}{\Rightarrow} = x^{2} + 2 x + 4 \\ u^{2} - 2 u + 4 \overset{u = x + 2}{\Rightarrow} x^{2} + 2 x + 1 \end{cases} \Rightarrow$ $\underset{i = u}{\prod^{m}} (u^{2} + 2 u + 4) = \underset{i + 2 = u}{\prod^{m + 2}} (u^{2} - 2 u + 4)$ ${\begin{cases} u - 2 \overset{u = x + 4}{\Rightarrow} x + 2 \\ u + 2 \overset{u = x}{\Rightarrow} x + 2 \end{cases} \Rightarrow \underset{i + 4 = u}{\prod^{m + 4}} u - 2 = \underset{i = u}{\prod^{m}} u + 2$ $\Rightarrow \underset{u = 3}{\prod^{\infty}} \frac{u^{3} - 8}{u^{3} + 8} = \underset{u = 3}{\prod^{\infty}} \frac{(u - 2) (u^{2} + 2 u + 4)}{(u + 2) (u^{2} - 2 u + 4)} =$ $\frac{1 \times 2 \times 3 \times 4}{7 \times 12} \times \frac{\underset{u = 7}{\prod^{\infty}} (u - 2) \times \underset{3}{\prod^{\infty}} (u^{2} + 2 u + 4)}{\underset{u = 3}{\prod^{\infty}} (u + 2) \times \underset{5}{\prod^{\infty}} (u^{2} - 2 u + 4)} =$ $\frac{2}{7}$
	Commented by amin96 last updated on 30/Jan/22

	$\frac{2}{7}$
	Answered by puissant last updated on 31/Jan/22

	$k = n + 2$ $\underset{n = 1}{\prod^{\infty}} \frac{{(n + 2)}^{3} - 2^{3}}{{(n + 2)}^{2} + 2^{3}} = \underset{k = 3}{\prod^{\infty}} \frac{k^{3} - 2^{3}}{k^{3} + 2^{3}}$ $= \underset{k = 3}{\prod^{\infty}} \frac{(k - 2) (k^{2} + 2 k + 4)}{(k + 2) (k^{2} - 2 k + 4)}$ $= \lim_{k \to \infty} \underset{t = 3}{\prod^{k}} (\frac{t - 2}{t + 2}) \times \lim_{k \to \infty} \underset{t = 3}{\prod^{k}} (\frac{t^{2} + 2 t + 4}{t^{2} - 2 t + 4})$ $= \lim_{k \to \infty} \frac{1}{5} ∙ \frac{2}{6} ∙ \frac{3}{7} ∙ \frac{4}{8} ∙ \frac{5}{9} ∙ . . . . ∙ \frac{k - 2}{k + 2} \times$ $\lim_{k \to \infty} \frac{19}{7} ∙ \frac{28}{12} ∙ \frac{39}{19} ∙ \frac{52}{28} ∙ . . . ∙ \frac{k^{2} + 3}{k^{2} - 4 k + 7} ∙ \frac{k^{2} + 2 k + 4}{k^{2} - 2 k + 4}$ $= 24 \lim_{k \to \infty} \frac{1}{(k - 1) k (k + 1) (k + 2)} \times$ $\frac{1}{7 \times 12} \lim_{k \to \infty} (k^{2} + 3) (k^{2} + 2 k + 4)$ $= \frac{24}{84} \lim_{k \to \infty} \frac{(k^{2} + 3) (k^{2} + 2 k + 4)}{(k - 1) k (k + 1) (k + 2)} = \frac{24}{84} .$ $\underset{n = 1}{\prod^{\infty}} \frac{{(n + 2)}^{3} - 8}{{(n + 2)}^{3} + 8} = \frac{2}{7} . . .$ $. . . . . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum