数列{an}中,a1=1,an,an+1是方程x2-(2n+1)x+1bn=0的两个根,则数列{bn}的前n项和Sn=( ) A.12n+1 B.1n+1 C.n2n+1 D.nn+1
问题描述:
数列{an}中,a1=1,an,an+1是方程x2-(2n+1)x+
=0的两个根,则数列{bn}的前n项和Sn=( )1 bn
A.
1 2n+1
B.
1 n+1
C.
n 2n+1
D.
n n+1
答
依题意,an+an+1=2n+1,
∴an+1+an+2=2(n+1)+1,
两式相减得:an+2-an=2,又a1=1,
∴a3=1+2=3,a5=5,…
∵an+an+1=2n+1,a1=1,
∴a2=3-1=2,a4=2+2=4,…
∴an=n;
又
=anan+1=n(n+1),1 bn
∴bn=
=1 n(n+1)
-1 n
,1 n+1
∴Sn=b1+b2+…+bn=(1-
)+(1 2
-1 2
)+…+(1 3
-1 n
)=1-1 n+1
=1 n+1
.n n+1
故选D.