Question

已知数列{an}的前n项和为sn，且满足sn=1/4（an+1）^2

1）求{an}通项公式
2）设bn=1/[anXa(n+1)]，则数列bn的前n项和Tn为

Answer 1

1) a1=1/4(a1+1)^2
a1=1
an=Sn-S(n-1)=1/4(an+1)^2-1/4[a(n-1)+1]^2
(an)^2-2an=[a(n-1)]^2+2a(n-1)
(an-1)^2=[a(n-1)+1]^2
an-1=±[a(n-1)+1]
an-a(n-1)=2,或an=-a(n-1)
因此，an=a1+(n-1)d=2n-1,或an=(-1)^(n-1)
2) bn=1/[anXa(n+1)]
Tn=1/(1*3)+1/(3*5)+1/(5*7)+...+1/[(2n-1)(2n+1)]
=1/2[1-1/3+1/3-1/5+1/5-1/7+....+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)
或Tn=1/(-1*1)+1/(-1*1)+...+1/[(-1)^(n-1)*(-1)^(n+1)]
=-1+(-1)+...+(-1)
=-n