{an}是首相为1,公差为2的等差数列,bn=1/ana(n+1)求数列{bn}的前n项和

An=1+(n-1)×2=2n-1
Bn=1/(AnA(n+1))=(1/d)×(A(n+1)-An)/(AnA(n+1))=(1/d)(1/An-1/A(n+1)
=1/2×(1/(2n-1)-1/(2n+1))
Tn=B1+B2+B3+……+Bn
=1/2×[(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+……+(1/(2n-1)-1/(2n+1)]
=1/2×(1-1/(2n+1))
=1/2×2n/(2n+1)
=n/(2n+1)