anonymous
  • anonymous
f(n) = integer nearest sqrt(n) 1/f(1) + 1/f(2) + ...........+1/f(9999) +1/f(10000) = ?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
|dw:1350239055071:dw|
anonymous
  • anonymous
That looks promising...
anonymous
  • anonymous
So when is f(n) = m?

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

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