设数列{an}的前n项和为Sn,a1=1,Sn+1=4an+2(n∈N+)(1)若bn=an+1-2an,求bn;(2)若cn=1an+1-2an,求{cn}的前6项和T6;(3)若dn=an2n,证

题目简介

设数列{an}的前n项和为Sn,a1=1,Sn+1=4an+2(n∈N+)(1)若bn=an+1-2an,求bn;(2)若cn=1an+1-2an,求{cn}的前6项和T6;(3)若dn=an2n,证

题目详情

设数列{an}的前n项和为Sn,a1=1,Sn+1=4an+2(n∈N+
(1)若bn=an+1-2an,求bn
(2)若cn=
1
an+1-2an
,求{cn}的前6项和T6
(3)若dn=
an
2n
,证明{dn}是等差数列.
题型:解答题难度:中档来源:不详

答案

解(1)∵a1=1,Sn+1=4an+2(n∈N+),
∴Sn+2=4an+1+2an+2=Sn+2-Sn+1=4(an+1-an),
∴an+2-2an+1=2(an+1-2an)
即bn+1=2bn
∴{bn}是公比为2的等比数列,且b1=a2-2a1
∵a1=1,a2+a1=S2
即a2+a1=4a1+2,
∴a2=3a1+2=5,
∴b1=5-2=3,
bn=3•2n-1
(2)∵cn=class="stub"1
an+1-2an
=class="stub"1
bn
=class="stub"1
3•2n-1

c1=class="stub"1
3•21-1
=class="stub"1
3
,∴cn=class="stub"1
3
•(class="stub"1
2
)n-1

∴{cn}是首项为class="stub"1
3
,公比为class="stub"1
2
的等比数列.
T6=
class="stub"1
3
[1-(class="stub"1
2
)
6
]
1-class="stub"1
2
=class="stub"2
3
(1-class="stub"1
64
)=class="stub"61
96

(3)∵dn=
an
2n
bn=3•2n-1

dn+1-dn=
an+1
2n+1
-
an
2n
=
an+1-2an
2n+1
=
bn
2n+1

dn+1-dn=
3•2n-1
2n+1
=class="stub"3
4

∴{dn}是等差数列.

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