问题
解答题
已知数列{an}是首项为1的等差数列,若a2+1,a3+1,a5成等比数列. (1)求数列{an}通项公式; (2)设bn=
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答案
(1)∵a2+1,a3+1,a5成等比数列.
∴(a3+1)2=a5•(a2+1)
即(2+2d)2=(1+4d)(2+d)
解可得,d=2,
∴an=1+2(n-1)=2n-1
(2)∵bn=
=1 anan+1
=1 (2n-1)(2n+1)
(1 2
-1 2n-1
)1 2n+1
∴sn=
(1-1 2
+1 3
-1 3
+…+1 5
-1 2n-1
)1 2n+1
=
(1-1 2
)=1 2n+1 n 2n+1