Question-185395

Question Number 185395 by mathlove last updated on 21/Jan/23

Answered by witcher3 last updated on 21/Jan/23

x^4 +(1/4)=(x^2 −x+(1/2))(x^2 +x+(1/2)) (2k+1)^2 −(2k+1)+(1/2)=(2k)^2 +2k+(1/2) (2k+1)^2 +(2k+1)+(1/2)=(2k)^2 +6k+2+(1/2) (2k+2)^2 −(2k+2)+(1/2)=4k^2 +6k+2+(1/2) ((Π_(k=0) ^9 ((2k+1)^2 +(1/4)))/(Π_(k=1) ^(10) ((2k)^2 +(1/4))))=((.Π_(k=0) ^9 ((2k)^2 +2k+(1/2))((2k)^2 +6k+2+(1/2)))/(Π_(k=0) ^9 ((2k)^2 −2k+(1/2))((2k)^2 +2k+(1/2)))) =Π_(m=0) ^9 ((((2k)^2 +6k+2+(1/2)))/((2k)^2 −2k+(1/2))) =2.((Π_(m=0) ^9 (4k^2 +6k+2+(1/2)))/(Π_(n=0) ^8 (4n^2 +6n+2+(1/2))))=2.(4.9^2 +6.9+2+(1/2)) =765

$$\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}=\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{2k}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\left(\mathrm{2k}\right)^{\mathrm{2}} +\mathrm{2k}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2k}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\left(\mathrm{2k}\right)^{\mathrm{2}} +\mathrm{6k}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{2k}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{2k}+\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{4k}^{\mathrm{2}} +\mathrm{6k}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{9}} {\prod}}\left(\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\right)}{\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{10}} {\prod}}\left(\left(\mathrm{2k}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\right)}=\frac{.\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{9}} {\prod}}\left(\left(\mathrm{2k}\right)^{\mathrm{2}} +\mathrm{2k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\left(\mathrm{2k}\right)^{\mathrm{2}} +\mathrm{6k}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{9}} {\prod}}\left(\left(\mathrm{2k}\right)^{\mathrm{2}} −\mathrm{2k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\left(\mathrm{2k}\right)^{\mathrm{2}} +\mathrm{2k}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\underset{\mathrm{m}=\mathrm{0}} {\overset{\mathrm{9}} {\prod}}\frac{\left(\left(\mathrm{2k}\right)^{\mathrm{2}} +\mathrm{6k}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\mathrm{2k}\right)^{\mathrm{2}} −\mathrm{2k}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{2}.\frac{\underset{\mathrm{m}=\mathrm{0}} {\overset{\mathrm{9}} {\prod}}\left(\mathrm{4k}^{\mathrm{2}} +\mathrm{6k}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\underset{\mathrm{n}=\mathrm{0}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{4n}^{\mathrm{2}} +\mathrm{6n}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\mathrm{2}.\left(\mathrm{4}.\mathrm{9}^{\mathrm{2}} +\mathrm{6}.\mathrm{9}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{765} \\ $$

Commented by mathlove last updated on 21/Jan/23

$${thanks}\:{dear} \\ $$

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