北师版高中同步学案数学选择性必修第二册精品课件 第1章 数列 培优课1 数列的通项公式问题——分层作业.ppt

第1章培优课1数列的通项公式问题

1234567891011121314151617A级必备知识基础练181.[探究点二]已知等差数列{an}的前n项和为Sn,且a5=5,a1+S11=67,则a3a10是{an}中的()A.第30项 B.第36项 C.第48项 D.第60项A解析设等差数列{an}的公差为d,由a5=5,得a1+4d=5①;由a1+S11=67,得12a1+d=67,即12a1+55d=67②.由①②解得a1=1,d=1,所以an=n.于是a3a10=3×10=30,而a30=30,故a3a10是{an}中的第30项.故选A.

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1234567891011121314151617184.[探究点三]已知等比数列{an}的前n项和为Sn=3n+1+t,则数列的通项公式an=.?2×3n解析∵在等比数列{an}中,前n项和Sn=3n+1+t,∴a1=S1=9+t,a2=S2-S1=18,a3=S3-S2=54,∴182=54(9+t),解得t=-3,∴a1=9+t=6,公比q=3,∴an=6×3n-1=2×3n.

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1234567891011121314151617186.[探究点二、三]数列{an}满足a1=1,Sn+1=4an+3.(1)求证:数列{an+1-2an}是等比数列;(2)求数列{an}的通项公式.

123456789101112131415161718(1)证明当n=1时,a1=1,S2=a1+a2=4a1+3,解得a2=6,当n≥2时,由Sn+1=4an+3可知Sn=4an-1+3,两式作差可得an+1=4an-4an-1,即an+1-2an=2(an-2an-1),又因为a2-2a1=4,所以an-2an-1≠0,所以数列{an+1-2an}是首项为4,公比为2的等比数列.

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1234567891011121314151617187.[探究点二、三]已知数列{an}的前n项和Sn满足an·Sn=(Sn-1)2.(1)证明:数列为等差数列;(2)求数列{an}的通项公式.

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123456789101112131415161718B级关键能力提升练8.已知在数列{an}中,a1=1,an+1=an+n,则数列{an}的通项公式为()C解析因为an+1=an+n,所以an=an-1+n-1(n≥2).又因为a1=1,则an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=(n-1)+(n-2)+…+1+1故选C.

1234567891011121314151617189.在数列{an}中,若a1=1,an+1=2an+3(n∈N+),则该数列的通项an=()A.2n+1-3 B.2n-3C.2n+1+3 D.2n+1-1A解析由an+1=2an+3,得an+1+3=2(an+3),又a1=1,∴a1+3=4≠0,∴数列{an+3}是以4为首项,以2为公比的等比数列,则an+3=4×2n-1,∴an=2n+1-3.故选A.

123456789101112131415161718A.a8 B.2+(n-1)lnnC.1+n+lnn D.2n+nlnnD

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12345678910111213141516171811.已知数列{an}满足a1=1,an=n(an+1-an),则数列{an}的通项公式为()A.an=2n-1 C.an=n2 D.an=nD

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12345678910111213141516171812.已知数列{an}的前n项和为Sn,a2=6,(n∈N+),则数列{an}的通项公式为()A.an=3n B.an=3nC.an=n+4 D.an=n2+2A

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12345678910111213141516171813.正项数列{an}满足anan+2=,n∈N+.若a5=9,a2a4=1,则a2的值为.?

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