问题
选择题
若数列{an}的前n项和Sn=n2（n∈N*），则
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答案
当n=1时，a1=s1=1
当n≥2时，an=Sn-Sn-1=n2-（n-1）2=2n-1
总之an=2n-1
∴
=1 anan+1
=1 (2n-1)(2n+1)
(1 2
-1 2n-1
)1 2n+1
∴
+1 a1a2
+…+1 a2a3 1 anan+1
=
(1-1 2
+1 3
-1 3
+1 5
-1 5
+…+1 7
-1 2n-1
)1 2n+1
=
(1-1 2
)1 2n+1
=n 2n+1
故选A
