(2012•德州一模)已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足Sn+2n=2an
(I)證明:數(shù)列{an+2}是等比數(shù)列,并求數(shù)列{an}的通項(xiàng)公式an
(Ⅱ)若數(shù)列{bn}滿足bn=log2(an+2),求數(shù)列{
1bn
}的前n項(xiàng)和Tn
分析:(I)由Sn+2n=2an,得Sn=2an-2n,由此利用構(gòu)造法能夠證明數(shù)列{an+2}是等比數(shù)列,并求出數(shù)列{an}的通項(xiàng)公式an
(Ⅱ)由an=2n+1-2,得
1
bn
=
1
n(n+1)
=
1
n
-
1
n+1
,由此利用錯(cuò)位相減法能夠求出數(shù)列{
1
bn
}的前n項(xiàng)和Tn
解答:(I)證明:由Sn+2n=2an,得Sn=2an-2n,
當(dāng)n∈N*時(shí),Sn=2an-2n,①
當(dāng)n=1時(shí),S1=2a1-2,則a1=2,
當(dāng)n≥2時(shí),Sn-1=2an-1-2(n-1),②
①-②,得an=2an-2an-1-2,
即an=2an-1+2,
∴an+2=2(an-1+2),
an+2
an-1+2
=2
,
∴{an+2}是以a1+2為首項(xiàng),以2為公比的等比數(shù)列.
an+2=4•2n-1,
an=2n+1-2
(Ⅱ)解:∵an=2n+1-2
∴bn=n(n+1),
1
bn
=
1
n(n+1)
=
1
n
-
1
n+1

Tn=
1
b1
+
1
b2
+…+
1
bn

=1-
1
2
+
1
2
-
1
3
+…+
1
n
-
1
n+1

=1-
1
n+1

=
n
n+1
點(diǎn)評(píng):本題考查等比數(shù)列的證明,考查數(shù)列的通項(xiàng)公式和前n項(xiàng)和的求法,解題時(shí)要認(rèn)真審題,注意構(gòu)造法和錯(cuò)位相減法的合理運(yùn)用.
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(2012•德州一模)對(duì)于直線m,n和平面α,β,γ,有如下四個(gè)命題:
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