问题
选择题
数列{an}中,a1=1,an,an+1是方程x2-(2n+1)x+
|
答案
由题意可得an+an+1=2n+1
∴an=n
∵anan+1=
∴bn=1 bn
=1 n(n+1)
-1 n 1 n+1
Sn=b1+b2+…+bn
=(1-
+1 2
-1 2
+…1 3
-1 n
)1 n+1
=1-
=1 n+1 n n+1
故选B.
数列{an}中,a1=1,an,an+1是方程x2-(2n+1)x+
|
由题意可得an+an+1=2n+1
∴an=n
∵anan+1=
∴bn=1 bn
=1 n(n+1)
-1 n 1 n+1
Sn=b1+b2+…+bn
=(1-
+1 2
-1 2
+…1 3
-1 n
)1 n+1
=1-
=1 n+1 n n+1
故选B.