(2012•黄浦区一模)已知a<b,且a2-a-6=0,b2-b-6=0,数列{an}、{bn}满足a1=1,a2=-6
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(2012•黄浦区一模)已知a<b,且a2-a-6=0,b2-b-6=0,数列{an}、{bn}满足a1=1,a2=-6a,a
证明(1)∵a<b,a2-a-6=0,b2-b-6=0,
∴a=-2,b=3,a2=12.
∵an+1=6an−9an−1(n≥2,n∈N*),bn=an+1−ban(n∈N*),
∴bn+1=an+2-3an+1
=6an+1-9an+1-3an+1
=3(an+1-3an)
=3bn (n∈N*).
又b1=a2-3a1=9,
∴数列{bn}是公比为3,首项为b1的等比数列.
(2)依据(1)可以,得bn=3n+1(n∈N*).
于是,有an+1-3an=3n+1(n∈N*),即
an+1
3n+1−
an
3n=1,(n∈N*).
因此,数列{
an
3n}是首项为
a1
31=
1
3,公差为1的等差数列.
故
an
3n=
1
3+(n−1)•1.
所以数列{an}的通项公式是an=(3n-2)•3n-1(n∈N*).
(3)用数学归纳法证明:cn +acn−1=
an
3n−2(n≥2,n∈N*)
(i)当n=2时,左边:cn+acn-1=c2-2c1=3,
右边:
an
3n−2=
(3×2−2)•32−1
3×2−2=3,
即左边=右边,所以当n=2时结论成立.
(ii)假设当n=k.(k≥2,k∈N*)时,结论成立,即ck +ack−1=
ak
3k−2(k≥2,k∈N*).
当n=k+1时,左边=ck+1+ack
=5ck-6ck-1-2ck
=3(ck-2ck-1)=3•
ak
3k−2=3k,
右边=
ak+1
3(k+1)−2=
∴a=-2,b=3,a2=12.
∵an+1=6an−9an−1(n≥2,n∈N*),bn=an+1−ban(n∈N*),
∴bn+1=an+2-3an+1
=6an+1-9an+1-3an+1
=3(an+1-3an)
=3bn (n∈N*).
又b1=a2-3a1=9,
∴数列{bn}是公比为3,首项为b1的等比数列.
(2)依据(1)可以,得bn=3n+1(n∈N*).
于是,有an+1-3an=3n+1(n∈N*),即
an+1
3n+1−
an
3n=1,(n∈N*).
因此,数列{
an
3n}是首项为
a1
31=
1
3,公差为1的等差数列.
故
an
3n=
1
3+(n−1)•1.
所以数列{an}的通项公式是an=(3n-2)•3n-1(n∈N*).
(3)用数学归纳法证明:cn +acn−1=
an
3n−2(n≥2,n∈N*)
(i)当n=2时,左边:cn+acn-1=c2-2c1=3,
右边:
an
3n−2=
(3×2−2)•32−1
3×2−2=3,
即左边=右边,所以当n=2时结论成立.
(ii)假设当n=k.(k≥2,k∈N*)时,结论成立,即ck +ack−1=
ak
3k−2(k≥2,k∈N*).
当n=k+1时,左边=ck+1+ack
=5ck-6ck-1-2ck
=3(ck-2ck-1)=3•
ak
3k−2=3k,
右边=
ak+1
3(k+1)−2=
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