## estudier 3 years ago f(n) = integer nearest sqrt(n) 1/f(1) + 1/f(2) + ...........+1/f(9999) +1/f(10000) = ?

1. mahmit2012

|dw:1350239055071:dw|

2. estudier

That looks promising...

3. estudier

So when is f(n) = m?

4. experimentX

*

5. estudier

Get rid of them nasty radicals as well.....

6. sara12345

f(n) would range from 1 to 100

7. sara12345

how to get rid of radical

8. estudier

f(n) = a when (a-1/2)^2 < n < (a+1/2)^2 (I'm using a)

9. sara12345

ok

10. estudier

Think I will leave this one up overnight in case people want to think more about it....

11. mukushla

and im watching :)

12. sara12345

does this involve integrals ?

13. estudier

Certainly not! (I'm not that fond of calculus)

14. sara12345

2a is the window for n until a increments

15. sara12345

is this right direction

16. estudier

Yes! Now u have the pattern, away u go....:-)

17. sara12345

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

18. estudier

Then the sum is ?

19. estudier

The last bit is just some counting (don't overcount)

20. sara12345

lol its easy

21. sara12345

2a.1/a a would cancel out ! 99*2 + 1 = 199

22. estudier

Yes, well done...

23. sara12345

nice exercise thnk you :)

24. estudier

yw

25. estudier

Since it has been solved I will close it after all....