已知等差数列{an}中，a1=1,公差d=2

a(n+1)-an=2
所以1/[an*a(n+1)]=(1/2)*[1/an-1/a(n+1)]
a(n+1)=a1+nd=1+2n
所以1/a1a2 +1/a2a3 +…+1/ana(n+1)
=1/2*[1/a1-1/a2+1/a2-1/a3+……+1/an-1/a(n+1)]
=1/2*[1/a1-1/a(n+1)]
=1/2*[1-1/(1+2n)]
=n/(1+2n)

1/a1a2 +1/a2a3 +…+1/ana(n+1)=17/35,
n/(1+2n)=17/35=17/(2*17+1)
所以n=17

引用：等差数列{an}，a1=1,公差d=2
解：-an+a(n+1))=2,
(1/an-1/a(n+1))=(-an+a(n+1))/(ana(n+1))=2/(ana(n+1))
于是：
1/a1a2 +1/a2a3 +…+1/ana(n+1)
=(1/a1-1/a2+1/a2-1/a3+...+1/an-1/a(n+1))/2
=(1/a1-1/a(n+1))/2
=(1-1/(1+2n))/2
=n/(1+2n)