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Topic: Which is larger? (Read 779 times) |
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pcbouhid
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Which is larger?
« on: Jan 2nd, 2006, 10:54am » |
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(143)Which is larger, 99^n + 100^n or 101^n, where n is a natural number?
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towr
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Re: Which is larger?
« Reply #1 on: Jan 2nd, 2006, 11:36am » |
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Going on the suggestion that it's the same for all n, obviously 99^n + 100^n would be greater, since it's greater when n is 0 or 1.
But possibly the reverse might be the case when n is great enough. [e] and for n > 48 that's in fact the case[/e]
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| « Last Edit: Jan 2nd, 2006, 12:05pm by towr » |
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Miles
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Re: Which is larger?
« Reply #2 on: Aug 24th, 2006, 2:43pm » |
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I don’t get the feeling that anyone’s waiting keenly for more on this, but I felt like setting out a method that could be generalised.
Dividing each expression by 100^n and taking the difference, the question is equivalent to determining the sign of f(n) = 0.99^n + 1 – 1.01^n.
f(0) = 1, but f(n) is dominated by the third term for large n, so f(n)-> –inf as n->inf. So f(n) must change sign.
Writing 0.99 as (1 – 0.01) and 1.01 as (1 + 0.01), expanding and cancelling
f(n) = 1 – 0.02n – n(n-1)(n-2) / 3 * 0.01^3 + terms in 0.01^5 etc
So very quickly from the first two terms, we see that if n=50, f(n) = (approx) 0.
With more care, we can show that f(48) > 0 and f(49) < 0. Anyone care to do that rigorously?
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