{an}的等差数列,公差d>0.S4=24,a2*a3=35 求：an、若bn=1/anan+1,求{bn}的前n项和Tn
问题描述：
{an}的等差数列,公差d>0.S4=24,a2*a3=35 求：an、若bn=1/anan+1,求{bn}的前n项和Tn
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答
4a1+(1+2+3)d=24
(a1+d)(21+2d)=35
解得
a1=3,d=2
an=3+2(n-1)=2n+1
bn=1/(an*a(n+1))=1/an - 1/a(n+1)
Tn=b1+b2+……+bn
=1/3-1/5+1/5-1/7+1/7-1/9+……+1/(2n+1)-1/(2n+3)=1/3-1/(2n+3)