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Question Number 180746 by Acem last updated on 16/Nov/22

Answered by MJS_new last updated on 16/Nov/22

$${(1+\frac{1}{200})}^{200}>2$$$$\mathrm{because}\mathrm{we}\mathrm{know}$$$${(1+\frac{1}{1})}^1=2$$$${(1+\frac{1}{n})}^n=\frac{{(n+1)}^n}{n^n}\mathrm{is}\mathrm{increasing}$$$$\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{1}{n})}^n=\mathrm{e}>2$$

Answered by mr W last updated on 17/Nov/22

$${(1+x)}^n=1+nx+\frac{n(n-1)x^2}{2}+...>1+nx$$$$withx=0.005andn=200:$$$$1.{005}^{200}>1+200\times 0.005=2✓$$$$even:$$$$1.{005}^{200}>1+200\times 0.005+\frac{1}{2}+(200\times 0.05)(200\times 0.05-0.005)$$$$=1+1+\frac{1\times 0.995}{2}>2.44$$$$even:$$$$1.{005}^{200}>1+1+\frac{1\times 0.995}{2}+\frac{1\times 0.995\times 0.99}{6}>2.66$$

Commented by manxsol last updated on 17/Nov/22

Commented by manxsol last updated on 17/Nov/22

$$thereasoningizfine???$$

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