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Is there some relatively nice way to show that $100^{100}-99^{99}>9.92\cdot 10^{199}$ by using only pencil and paper?

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did you try writing $9.92=10-0.08$ and then cancelling $10^{200}$? Did it help? – Alex Jan 28 '14 at 13:46
No but how can I cancel the number $10^{200}$? – selfstudying Jan 28 '14 at 13:51
$100^{100}=10^{200}$ – Alex Jan 28 '14 at 13:54
up vote 7 down vote accepted

By Bernoulli inequality, $$\frac{100^{99}}{99^{99}}=\left(1+\frac1{99}\right)^{99}>2$$ (in fact it is $\approx e$), so that $99^{99}<\frac12100^{99}=0.05\cdot 10^{199}$ and finally $$ 100^{100}-99^{99}>(10-0.05)10^{199}=9.95\cdot 10^{199}.$$

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Start with $99^{99} = (100-1)^{99}=100^{100}(1-\frac{1}{100})^{99}$. Then expand this using the binomial approximation. So have terms like 1-99/100 + 98*99/(100*100 * 2!) etc. Feel there is a neater way but can't see it.

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