Question

bn=2n+1,Tn为bn的和,求1/T1+1/T2+1/T3+……1/Tn

Answer 1

Tn=b1+b2+……bn

=(2×1+1)+(2×2+1)+……+(2n+1)

=2(1+2+……+n)+n

=2×n(n+1)/2+n

=n(n+1)+n

=n(n+2)

所以1/Tn=1/[n(n+2)]=1/2 [1/n-1/(n+2)]

1/T1+1/T2+1/T3+……1/Tn

=1/2 [1-1/3+1/2-1/4+1/3-1/5+……+1/n-1/(n+2)]

=1/2 (1+1/2-1/(n+1)-1/(n+2))

=3/4-1/[2(n+1)]-1/[2(n+2)]