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

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embizze | High School Teacher | (Level 1) Educator Emeritus

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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`



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