解题思路:(1)由题设条件可知
S
1
=
a
1
+3
a
1
2
,
a
1
=2
a
1
,即
a
1
=0
.由此能够解得a=0.
(2)由题意可知,
S
n
=
n
a
n
2
,2
S
n
=n
a
n
(n∈
N
*
)
.所以2Sn-1=(n-1)an-1(n≥2).由此可知数列{an}的通项公式an=2(n-1)(n∈N*).
(3)由题设条件知
S
n
=
n
a
n
2
=n(n−1)(n∈
N
*
)
.由此可知Tn=t1+t2+…+tn=
3−
2
n+1
−
2
n+2
<3(n∈
N
*
)
.从而求得数列{Tn}的上渐近值是3.
(1)∵a1=a,a2=2,Sn=
n(an+3a1)
2(n∈N*),∴S1=
a1+3a1
2,a1=2a1,即a1=0.(2分)∴a=0.(3分)
(2)由(1)可知,Sn=
nan
2,2Sn=nan(n∈N*).
∴2Sn-1=(n-1)an-1(n≥2).
∴2(Sn-Sn-1)=nan-(n-1)an-1,2an=nan-(n-1)an-1,(n-2)an=(n-1)an-1.(5分)
∴
an
n−1=
an−1
n−2(n≥3,n∈N*).(6分)
因此,
an
n−1=
an−1
n−2═
a2
1,an=2(n−1)(n≥2).(8分)
又a1=0,∴数列{an}的通项公式an=2(n-1)(n∈N*).(10分)
(3)由(2)有,Sn=
nan
2=n(n−1)(n∈N*).于是,tn=
Sn+2
Sn+1+
Sn+1
Sn+2−2
=
(n+2)(n+1)
(n+1)n+
(n+1)n
(n+2)(n+1)−2
=
2
n−
2
n+2(n∈N*).(12分)
∴Tn=t1+t2+…+tn
=(
2
1−
2
3)+(
2
2−
点评:
本题考点: 数列递推式;极限及其运算;数列的求和.
考点点评: 本题考查数列的综合运用,解题时要注意计算能力的培养.