用数学归纳法证明
(1)当n=1时,a1=1/4,f(n)=1-a1=1-1/4=3/4=(1+2)/2*(1+1),成立
(2)假设当n=k（k>1的正整数）时,结论成立,即f(k)=(1-a1)(1-a2)…(1-ak)=(k+2)/2(k+1)
则,当n=(k+1)时,a(k+1)=1/(k+2)^2,
f(k+1)=(1-a1)(1-a2)…(1-ak)*[1-a(k+1)]
=(k+2)/2(k+1) * [1-a(k+1)] 将a(k+1)带入化简,1-a(k+1)=(k+1)(k+3)/(k+2)^2
=(k+2)/2(k+1) * (k+1)(k+3)/(k+2)^2 约分
=(k+3)/2(k+2)
=[(k+1)+2]/2[(k+1)+1]
则对任意正整数,都有f(n)=(n+2)/2(n+1)
得证