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f(n) = integer nearest sqrt(n) 1/f(1) + 1/f(2) + ...........+1/f(9999) +1/f(10000) = ?

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That looks promising...
So when is f(n) = m?

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Other answers:

Get rid of them nasty radicals as well.....
f(n) would range from 1 to 100
how to get rid of radical
f(n) = a when (a-1/2)^2 < n < (a+1/2)^2 (I'm using a)
Think I will leave this one up overnight in case people want to think more about it....
and im watching :)
does this involve integrals ?
Certainly not! (I'm not that fond of calculus)
2a is the window for n until a increments
is this right direction
Yes! Now u have the pattern, away u go....:-)
so f(n) = a a = 1 ; 2.1 times a = 2 ; 2.2 times .... a = 99; 2.99 times a = 100; 2.100/2 times
Then the sum is ?
The last bit is just some counting (don't overcount)
lol its easy
2a.1/a a would cancel out ! 99*2 + 1 = 199
Yes, well done...
nice exercise thnk you :)
Since it has been solved I will close it after all....

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