Question

数列An,a1=1,n≥2时,其前n项和Sn满足Sn²=an(Sn-0.5) ①求Sn表达式②设

Answer 1

已知数列{an}中a1=1,当n≥2时,其前n项和Sn满足Sn²=an(Sn-1/2)

(1)求Sn (2)设bn=Sn/(2n+1),求{bn}的前n项和Tn

【解】(1)Sn²=an(Sn-1/2)

Sn²=(Sn-S(n-1))(Sn-1/2) (n>=2)

S(n-1)Sn=1/2(S(n-1)-Sn)

变形得

1/Sn-1/S(n-1)=2 (n>=2)

设{1/Sn}为Cn

Cn-C(n-1)=2 (n>=2)

Cn为等差数列,d=2,b1=1/S1=1

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

Sn=1/(2n-1)

当n=1时

1/S1=1/a1=1也适合.

(2)

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

Tn=1/2(1-1/3+1/3-1/5+…+1/(2n-3)-1/(2n-1)+1/(2n-1)-1/(2n+1))

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

=n/(2n+1)