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|dw:1350239055071:dw|

That looks promising...

So when is f(n) = m?

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)

ok

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 :)

yw

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