设数列{an}是公差为d（d＞0）的等差数列,Sn为{an}的前n项和,已知S4=24,a2乘a3=35,（1）求数列{an}...设数列{an}是公差为d（d＞0）的等差数列,Sn为{an}的前n项和,已知S4=24,a2乘a3=35,（1）求数列{an}的通项公式（2）若bn=1／anan+1,且{bn}的前项和为Tn,求Tn
网友回答
(1)S4=2(a2+a3),a2+a3=12
a2a3=35
t^2-12t+35=0
a2=5,a3=7
an=2n+1
(2)1/an(an+1)=(1/2)[1/(2n+1)-1/(2n+3)]
Tn=1/2(1/3-1/5+1/5-1/7+...-1/(2n+1)+1/(2n+1)-1/(2n+3))
=n/[3(2n+3)]
======以下答案可供参考======
供参考答案1:
(1)S4=2(a2+a3),a2+a3=12
a2a3=35
t^2-12t+35=0
a2=5,a3=7
an=2n+1
(2)1/an(an+1)=(1/2)[1/(2n+1)-1/(2n+3)]
Tn=1/2(1/3-1/5+1/5-1/7+...-1/(2n+1)+1/(2n+1)-1/(2n+3))
=n/[3(2n+3)]