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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

|dw:1350239055071:dw|
That looks promising...
So when is f(n) = m?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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)
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....

Not the answer you are looking for?

Search for more explanations.

Ask your own question