由f(1)+f(2)+……+f(n-1)+f(n)=n^2 × f(n)
故f(1)+f(2)+……+f(n-1) =(n-1)^2 × f(n-1)
两式相减可得
f(n)=n^2 × f(n) - (n-1)^2 × f(n-1)
即(n^2-1)× f(n) = (n-1)^2 × f(n-1)
所以f(n)=(n-1)^2 × f(n-1) / (n^2-1)
= f(n-1) (n-1) / (n+1)
故递推得
f(2011)=f(2010)× 2010/2012
=f(2009)× (2010×2009)/(2012×2011)
……
=f(1)×(2010×2009×2008……×3×2×1)/(2012×2011×2010×……×5×4×3)
=f(1)×2/(2012×2011)
=2011×2/(2012×2011)
=1/1006