Question

数列题.已知数列{An}的前n项和为Sn,且Sn=n^2 +n,数列{bn}满足bn=1/AnA(n+1) ,Tn是数列

Answer 1

An=Sn-S(n-1)

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

=n^2+n-n^2+2n-1-n+1

=2n

A(n+1)=2(n+1)

bn=1/4 (1/n-1/(n+1)

Tn=1/4(1/1-1/2+1/2-1/3+.+1/n-1/(n+1))

=1/4(1-1/(n+1))

T9=1/4(1-1/10)=1/4 *9/10=9/40