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Not a terribly hard one, but one I like a lot.
Find the product:
inf
PI 10^(2^-n)
n=0
Submitted by Dave E |
Problem Moderated by: Graeme |
| Problem Solution |
the product is 10^0*10^.5*10^.25....
By rules of multiplication
10^(1+.5+.25+...2^-n)
The exponent becomes
inf
<sum> 1/2^n
n=0
This is a geometric series whose sum is 1/1-.5 = 2
Thus the product
inf
PI 10^(2^-n) = 10^2 = 100
n=0 |
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