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A function f is defined for all positive integers abd f(1)=2011. If for every n>1, f(1)+f(2)+f(3)+...+f(n)=n 2 .f(n), then f(2011)=? a)1/1004 b)1/2008 c)1/1006 d)1/2012 Plz explain...

A function f is defined for all positive integers abd f(1)=2011.


If for every n>1, f(1)+f(2)+f(3)+...+f(n)=n2.f(n), then f(2011)=?


a)1/1004


b)1/2008


c)1/1006


d)1/2012


 


Plz explain...

Grade:12

1 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
8 years ago
Do little re-arranging


Now its easy to solve the recurrence relation


Arun Kumar
IIT Delhi
Askiitians Faculty

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