问题
解答题
已知数列{an}为等差数列,{an}的前n项和为Sn,a1+a3=
(1)求数列{an}的通项公式; (2)若数列{bn}满足anbn=
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答案
(1)由a1+a3=
,S5=5,得3 2 2a1+2d= 3 2 5a1+
d=55×4 2
解得a1=
,d=1 2
.1 4
∴an=
.n+1 4
(2)∵an=
,∴bn=n+1 4
,1 n+1
∴bnbn+1=
=1 (n+1)(n+2)
-1 n+1
,1 n+2
∴Tn=b1b2+b2b3+b3b4+…+bnbn+1=(
-1 2
)+(1 3
-1 3
)+…+(1 4
-1 n+1
)=1 n+2
-1 2
=1 n+2
.n 2(n+2)