why-200-lt-100-200-

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

$${why}\:\:\:“\:\mathrm{200}!<\mathrm{100}^{\mathrm{200}} \:''\:? \\ $$

Commented by mehdee42 last updated on 17/May/23

99×101<100^2 98×102<100^2 : 2×198<100^2 1×199×100×200<100^4 ×<×⇒200!<100^(200)

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

Answered by Frix last updated on 16/May/23

Test it f(n)=(2n)!−n^(2n) f(1)=1 f(2)=8 f(3)=−9 f(4)=−25216 f(5)=−6136825 ⇒ (2n)!<n^(2n) ∀n≥3 200!<10^(375) 100^(200) =10^(400)

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

Answered by manxsol last updated on 16/May/23

((1×2×3...................200)/(100×100×100.......100))<1 0.01×0.02..... 1×1.01× 1.99× 2 (200term) 2×0.01×1.99×0.02...1×1.01 )(100 terms)<1

$$\frac{\mathrm{1}×\mathrm{2}×\mathrm{3}……………….\mathrm{200}}{\mathrm{100}×\mathrm{100}×\mathrm{100}…….\mathrm{100}}<\mathrm{1} \\ $$$$\mathrm{0}.\mathrm{01}×\mathrm{0}.\mathrm{02}…..\:\:\mathrm{1}×\mathrm{1}.\mathrm{01}×\:\:\:\:\:\:\:\mathrm{1}.\mathrm{99}×\:\:\mathrm{2}\:\left(\mathrm{200}{term}\right) \\ $$$$\left.\mathrm{2}×\mathrm{0}.\mathrm{01}×\mathrm{1}.\mathrm{99}×\mathrm{0}.\mathrm{02}…\mathrm{1}×\mathrm{1}.\mathrm{01}\:\right)\left(\mathrm{100}\:{terms}\right)<\mathrm{1} \\ $$

Answered by manxsol last updated on 17/May/23

((1+2+3......199)/(199))≫^(199) (√(1.2.3...199)) ((199×200)/(2(199)))≫^(199) (√( 199!)) 100^(199) ≫199! 200×100^(199) ≫200×199! 2×100^(200) ≫200!

$$\frac{\mathrm{1}+\mathrm{2}+\mathrm{3}……\mathrm{199}}{\mathrm{199}}\gg^{\mathrm{199}} \sqrt{\mathrm{1}.\mathrm{2}.\mathrm{3}…\mathrm{199}} \\ $$$$\frac{\mathrm{199}×\mathrm{200}}{\mathrm{2}\left(\mathrm{199}\right)}\gg^{\mathrm{199}} \sqrt{\:\mathrm{199}!} \\ $$$$\mathrm{100}^{\mathrm{199}} \gg\mathrm{199}! \\ $$$$\mathrm{200}×\mathrm{100}^{\mathrm{199}} \gg\mathrm{200}×\mathrm{199}! \\ $$$$\mathrm{2}×\mathrm{100}^{\mathrm{200}} \gg\mathrm{200}! \\ $$

Commented by manxsol last updated on 17/May/23

$${I}\:{think}\:{it}\:{is}\:{the}\:{solution}. \\ $$$$\:{something}\:{missing}.{Check}\:{it}\:{out} \\ $$

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