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
人气:185 ℃ 时间:2019-08-20 06:48:10
解答
(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)]
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