设{an}为公差大于0的等差数列,Sn为数列{an}的前n项的和.已知S4=24,a2a3=35. (Ⅰ)求数列{an}的通项公式an; (Ⅱ)若bn=1/anan+1,求{bn}的前n项和Tn.
问题描述:
设{an}为公差大于0的等差数列,Sn为数列{an}的前n项的和.已知S4=24,a2a3=35.
(Ⅰ)求数列{an}的通项公式an;
(Ⅱ)若bn=
,求{bn}的前n项和Tn. 1
anan+1
答
(I)∵S4=
=2(a2+a3)=24,4(a1+a4) 2
由
解得a2=5,a3=7,或a2=7,a3=5,(4分),
a2+a3=12
a2a3=35
∵d>0,
∴a2=5,a3=7,
于是d=a3-a2=2,a1=3,(6分)
∴an=3+2(n-1)=2n+1(18分)
(II)bn=
=1 (2n+1)(2n+3)
(1 2
-1 2n+1
)(10分)1 2n+3
∴Tn
[(1 2
-1 3
)+(1 5
-1 5
)+…+(1 7
-1 2n+1
)]=1 2n+3
(12分)n 6n+9