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

Answered by witcher3 last updated on 21/Jan/23

$${\mathrm{x}}^4+\frac{1}{4}=({\mathrm{x}}^2-\mathrm{x}+\frac{1}{2})({\mathrm{x}}^2+\mathrm{x}+\frac{1}{2})$$$${(2\mathrm{k}+1)}^2-(2\mathrm{k}+1)+\frac{1}{2}={\left(2\mathrm{k}\right)}^2+2\mathrm{k}+\frac{1}{2}$$$${(2\mathrm{k}+1)}^2+(2\mathrm{k}+1)+\frac{1}{2}={\left(2\mathrm{k}\right)}^2+6\mathrm{k}+2+\frac{1}{2}$$$${(2\mathrm{k}+2)}^2-(2\mathrm{k}+2)+\frac{1}{2}={4\mathrm{k}}^2+6\mathrm{k}+2+\frac{1}{2}$$$$\frac{\underset{\mathrm{k}=0}{\prod ^9}({(2\mathrm{k}+1)}^2+\frac{1}{4})}{\underset{\mathrm{k}=1}{\prod ^{10}}({\left(2\mathrm{k}\right)}^2+\frac{1}{4})}=\frac{.\underset{\mathrm{k}=0}{\prod ^9}({\left(2\mathrm{k}\right)}^2+2\mathrm{k}+\frac{1}{2})({\left(2\mathrm{k}\right)}^2+6\mathrm{k}+2+\frac{1}{2})}{\underset{\mathrm{k}=0}{\prod ^9}({\left(2\mathrm{k}\right)}^2-2\mathrm{k}+\frac{1}{2})({\left(2\mathrm{k}\right)}^2+2\mathrm{k}+\frac{1}{2})}$$$$=\underset{\mathrm{m}=0}{\prod ^9}\frac{({\left(2\mathrm{k}\right)}^2+6\mathrm{k}+2+\frac{1}{2})}{{\left(2\mathrm{k}\right)}^2-2\mathrm{k}+\frac{1}{2}}$$$$=2.\frac{\underset{\mathrm{m}=0}{\prod ^9}({4\mathrm{k}}^2+6\mathrm{k}+2+\frac{1}{2})}{\underset{\mathrm{n}=0}{\prod ^8}({4\mathrm{n}}^2+6\mathrm{n}+2+\frac{1}{2})}=2.(4.9^2+6.9+2+\frac{1}{2})$$$$=765$$

Commented by mathlove last updated on 21/Jan/23

$$thanksdear$$

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