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Question Number 192374 by mehdee42 last updated on 15/May/23

$$why\mathrm{♮}200!<{100}^{200}\epsilon ?$$

Commented bymehdee42 last updated on 17/May/23

$$99\times 101<{100}^2$$ $$98\times 102<{100}^2$$ $$:$$ $$2\times 198<{100}^2$$ $$1\times 199\times 100\times 200<{100}^4$$ $$\times <\times \Rightarrow 200!<{100}^{200}$$

Answered by Frix last updated on 16/May/23

$$\mathrm{Test}\mathrm{it}$$ $$f\left(n\right)=\left(2n\right)!-n^{2n}$$ $$f\left(1\right)=1$$ $$f\left(2\right)=8$$ $$f\left(3\right)=-9$$ $$f\left(4\right)=-25216$$ $$f\left(5\right)=-6136825$$ $$\Rightarrow \left(2n\right)!<n^{2n}\mathrm{\forall }n⩾3$$ $$$$ $$200!<{10}^{375}$$ $${100}^{200}={10}^{400}$$

Answered by manxsol last updated on 16/May/23

$$\frac{1\times 2\times 3...................200}{100\times 100\times 100.......100}<1$$ $$0.01\times 0.02.....1\times 1.01\times 1.99\times 2\left(200term\right)$$ $$2\times 0.01\times 1.99\times 0.02...1\times 1.01)\left(100terms\right)<1$$

Answered by manxsol last updated on 17/May/23

$$\frac{1+2+3......199}{199}{\gg }^{199}\sqrt{1.2.3...199}$$ $$\frac{199\times 200}{2\left(199\right)}{\gg }^{199}\sqrt{199!}$$ $${100}^{199}\gg 199!$$ $$200\times {100}^{199}\gg 200\times 199!$$ $$2\times {100}^{200}\gg 200!$$

Commented bymanxsol last updated on 17/May/23

$$Ithinkitisthesolution.$$ $$somethingmissing.Checkitout$$

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