解题思路:(1)由Sn+1=4an+2,得an+1=Sn+1-Sn=(4an+2)-(4an-1+2)(n≥2),由此能求出λ=2.
(2)由已知得bn=an+1-2an=4•2n-1,将an+1-2an=4•2n-1两边同除以2n+1,得
a
n+1
2
n+1
−
a
n
2
n
=4•2-2=1,由此能求出cn=n.
(3)由
c
n
=
a
n
2
n
=n,得
a
n
=n•
2
n
,从而dn=
2
n
n
−
2
n+1
n+1
,由此能求出
T
n
=2−
2
n+1
n+1
.
(1)由Sn+1=4an+2,得an+1=Sn+1-Sn=(4an+2)-(4an-1+2)(n≥2)
∴an+1-2an=2an-4an-1=2(an-2an-1)
故数列{an+1-2an} 是以a2-2a1为首项,2为公比的等比数列,
∵bn=an+1-λan,数列{bn}为等比数列,
∴λ=2.
(2)由a1=2,a1+a2=S2=4a1+2,
∴a2=8,
∴bn=an+1-2an=4•2n-1,
将an+1-2an=4•2n-1两边同除以2n+1,
得
an+1
2n+1−
an
2n=4•2-2=1,即cn+1-cn=1,
故{cn}是以c1=
a1
2=1为首项,1为公差的等差数列,
∴cn=n.
(3)∵cn=
an
2n=n,∴an=n•2n,
∴dn=([1
2log2
an/n]-[1
log2
an+1/n+1])•2n+1
=([1/2n−
1
n+1])•2n+1
=
2n
n−
2n+1
n+1,
∴Tn=2-
22
2+
22
2−
23
3+…+
2n
n−
2n+1
n+1
=2-
2n+1
n+1.
∴Tn=2−
2n+1
点评:
本题考点: 数列的求和;数列递推式.
考点点评: 本题考查实数值的求法,考查数列的通项公式的求法,考查数列的前n项和的求法,解题时要认真审题,注意裂项求和法的合理运用.