问题 解答题
已知向量
OP
=(x,y),
OQ
=(y,2)
,曲线C上的点满足:
OP
OQ
=2x
.点M(xk,xk+1)在曲线C上,且xk≠0,x1=1,数列{an}满足:ak=
1
xk
,(k,n∈N+)

(1)求数列{an}通项公式;
(2)若数列{bn}满足bn=7-2an,求数列{|bn|}的前n项和Tn
答案

(1)由题意可得xy+2y=2x,∴曲线C的方程为y=

2x
x+2
(x≠-2).

∵点M(xk,xk+1)在曲线C上,且xk≠0,∴xk+1=

2xk
xk+2

1
xk+1
=
1
xk
+
1
2

ak+1=ak+

1
2
,a1=1.

∴数列{an}是等差数列,

an=1+(n-1)×

1
2
=
n+1
2

(2)bn=7-2an=6-n.

当n≤6时,Tn=

n(5+6-n)
2
=
n(11-n)
2

当n>6时,Tn=15+

1
2
(n-6)(1+n-6)=
1
2
(n2-11n+60)

Tn=

n(11-n)
2
,n≤6
n2-11n+60
2
,n>6

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