在数列{an}中,已知a1=1,a(n+1)=2an/(an+2),求数列{anan+1}的前n项和

1/a(n+1)=an+2/2an=1/2+1/an
所以,{1/an}是公差为1/2的等差数列
1/an=1/a1+(n-1)*1/2=(n+1)/2
an=2/(n+1)
a(n+1)=2/(n+3)
ana(n+1)=4/(n+1)(n+3)=2(1/(n+1)-1/(n+3))
数列{anan+1}的前n项和
=2(1/(n+1)-1/(n+3))+2(1/(n-1)-1/(n+1))+...+2(1/2-1/4)
=2(1/2-1/(n+3))
=1-2/(n+3)
=(n+1)/(n+3)