Question

定义在正整数集上的函数f(x)满足 f(1)=2011且f(1)+f(2)+……+f(n)=n平方f(n)（n大于等于1

Answer 1

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