函数f(x)定义在正整数集上,且满足f(1)=2012和f(1)+f(2)+……+f(n)=n2f(n),则f(2011

f(1)+f(2)+……+f(n)=n²f(n)
f(1)+f(2)+……+f(n-1)=（n-1）²f(n-1) n≥2
两式相减
f(n)=n²f(n)-(n-1)²f(n-1)
(n²-1)f(n)=(n-1)²f(n-1) n≥2
(n+1)f(n)=(n-1)f(n-1)
f(n)/f(n-1)=(n-1)/(n+1)
f(2011)
=[f(2011)/f(2010)]*[f(2010)/f(2009)]*[f(2009)/f(2008)]*.*[f(3)/f(2)]*[f(2)/f(1)] *f(1)
=(2010/2012)*(2009/2011)*(2008/2010)*.*(2/4)*(1/3)*2012
=(1*2/2011*2012)*2012
=2/2011