f(1) = 2012  f(1) + f(2) + · · · + f(n) = n^2f(n); ∀n ∈ N.calculate the exactly value for-> f(2012)

Asked on by josanuz

1 Answer | Add Yours

embizze's profile pic

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

Given `f(1)=2012` and `f(n)=f(1)+f(2)+***+f(n)=n^2f(n)` find the exact value of `f(2012)` :

`f(1)=2012` is given

`f(2): f(1)+f(2)=4f(2) ==> 2012=3f(2)==>f(2)=2012/3`


Following the same procedure we find:




Noticing the denominators are all triangular numbers (see link: triangular numbers) we have:


So `f(2012)=2012/((2012(2013))/2)=2/2013`


Thus the required answer is `2/2013`



We’ve answered 319,829 questions. We can answer yours, too.

Ask a question