请问1/2an+1=1/2an+1是不是an=a(n+1)*(1+2an)?如果是的话,就是如下解决方法：an=a(n+1)*(1+2an)=a(n+1)+ 2a(n+1)*an,两端同除以a(n+1)*an,得到 1/a(n+1) = 1/an + 2 故数列1/an是等差数列,且1/a1=1,所以1/an=(2n-1),an= 1/(2n-1).因此a1a2+a2a3+……+anan+1 = 1/(1*3)+1/(3*5)+...+1/((2n-1)(2n+1)) =1/2 * [1/1-1/3 + 1/3-1/5 +...+1/(2n-1)-1/(2n+1)] =1/2 * [1-1/(2n+1)]=n/(2n+1),若a1a2+a2a3+……+anan+1>16/33,那么就有n/(2n+1)>16/33,所以33n>16(2n+1),求得n>16.