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

*print*Print*list*Cite

### 1 Answer

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`

`f(3):f(1)+f(2)+f(3)=9f(3)=>2012+2012/3=8f(3)=>f(3)=2012/6`

Following the same procedure we find:

`f(4)=2012/10`

`f(5)=2012/15`

`f(6)=2012/21`

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

`f(n)=2012/((n(n+1))/2)`

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

------------------------------------------

**Thus the required answer is** `2/2013`

------------------------------------------

**Sources:**