在数列an中,a1=1/2,点(an,an+1)在直线y=x+1/2上.记bn=1/anan+1,求数列bn的前n项和Tn

在数列an中,a1=1/2,点(an,an+1)在直线y=x+1/2上.记bn=1/anan+1,求数列bn的前n项和Tn

你确定是bn=1/anan+1，而不是bn=1/an +1 ?

点(an，an+1)在直线y=x+1/2上
a(n+1) = an + 1/2
a(n+1) -an = 1/2
an - a1 = (n-1)/2
an = n
bn=1/(an.a(n+1))
= 1/[n(n+1)]
=1/n -1/(n+1)
Tn =b1+b2+...+bn
=1- 1/(n+1)
=n/(n+1)