1人教A高中数学必修5同步训练1.设数列{(-1)n-1·n}的前n项和为Sn,则S2011等于()A.-2011B.-1006C.2011D.1006答案:D2.已知数列{1nn+1}的前n项和为Sn,则S9等于()A.910B.710C.109D.107答案:A3.数列{an}的通项公式an=1n+n+1,若前n项的和为10,则项数n为__________.答案:1204.求数列112,314,518,…,[(2n-1)+12n]的前n项和.解:Sn=112+314+518+…+[(2n-1)+12n]=(1+3+5+…+2n-1)+(12+14+18+…+12n)=1+2n-1·n2+12[1-12n]1-12=n2+1-12n.一、选择题1.在等差数列{an}中,已知a1=2,a9=10,则前9项和S9=()A.45B.52C.108D.54答案:D2.已知数列{an}的前n项和Sn=1-5+9-13+17-21+…+(-1)n-1(4n-3),则S15=()A.-29B.29C.30D.-30解析:选B.S15=1-5+9-13+…+57=-4×7+57=29.3.数列9,99,999,9999,…,的前n项和等于()A.10n-1B.1010n-19-nC.109(10n-1)D.109(10n-1)+n解析:选B.an=10n-1,∴Sn=a1+a2+…+an=(10-1)+(102-1)+…+(10n-1)=(10+102+…+10n)-n=1010n-19-n.24.已知数列{an}为等比数列,Sn是它的前n项和,若a2·a3=2a1,且a4与2a7的等差中项为54,则S5=()A.35B.33C.31D.29解析:选C.设公比为q(q≠0),则由a2·a3=2a1知a1q3=2,∴a4=2.又a4+2a7=52,∴a7=14.∴a1=16,q=12.∴S5=a11-q51-q=16[1-125]1-12=31.5.设等差数列{an}的前n项和为Sn,若a1=-11,a4+a6=-6,则当Sn取最小值时,n等于()A.6B.7C.8D.9解析:选A.设等差数列的公差为d,则由a4+a6=-6得2a5=-6,∴a5=-3.又∵a1=-11,∴-3=-11+4d,∴d=2,∴Sn=-11n+nn-12×2=n2-12n=(n-6)2-36,故当n=6时Sn取最小值,故选A.6.已知数列{an}:12,13+23,14+24+34,15+25+35+45,…,那么数列{bn}={1anan+1}前n项的和为()A.4(1-1n+1)B.4(12-1n+1)C.1-1n+1D.12-1n+1解析:选A.∵an=1+2+3+…+nn+1=nn+12n+1=n2,∴bn=1anan+1=4nn+1=4(1n-1n+1).∴Sn=4(1-1n+1).二、填空题7.已知an=n+13n,则数列{an}的前n项和Sn=__________.解析:Sn=(1+2+…+n)+(13+132+…+13n)=12(n2+n+1-13n).答案:12(n2+n+1-13n)8.若数列{an}的通项公式an=1n2+3n+2,则数列的前n项和Sn=__________.解析:an=1n2+3n+2=1n+1n+2=1n+1-1n+2,3Sn=(12-13)+(13-14)+…+(1n+1-1n+2)=12-1n+2=n2n+4.答案:n2n+49.已知数列{an}中,an=2n-1n为正奇数,2n-1n为正偶数,则a9=________(用数字作答),设数列{an}的前n项和为Sn,则S9=________(用数字作答).解析:a9=29-1=256.S9=(a1+a3+a5+a7+a9)+(a2+a4+a6+a8)=1-451-4+4×3+152=377.答案:256377三、解答题10.已知数列{an}的通项an=2·3n,求由其奇数项所组成的数列的前n项和Sn.解:由an=2·3n得an+1an=2·3n+12·3n=3,又a1=6,∴{an}是等比数列,其公比为q=3,首项a1=6,∴{an}的奇数项也成等比数列,公比为q2=9,首项为a1=6,∴Sn=61-9n1-9=34(9n-1).11.已知{an}是首项为19,公差为-2的等差数列,Sn为{an}的前n项和.(1)求通项an及Sn;(2)设{bn-an}是首项为1,公比为3的等比数列,求数列{bn}的通项公式及前n项和Tn.解:(1)∵{an}是首项为a1=19,公差为d=-2的等差数列,∴an=19-2(n-1)=21-2n,Sn=19n+12n(n-1)×(-2)=20n-n2.(2)由题意得bn-an=3n-1,即bn=an+3n-1,∴bn=3n-1-2n+21,Tn=Sn+(1+3+…+3n-1)=-n2+20n+3n-12.12.在数列{an}中,a1=1,an+1=2an+2n.(1)设bn=an2n-1,证明:数列{bn}是等差数列;(2)求数列{an}的前n项和Sn.解:(1)证明:由an+1=2an+2n,两边同除以2n,得an+12n=an2n-1+1.∴an+12n-an2n-1=1,即bn+1-bn=1,∴{bn}为等差数列.(2)由第(1)问得,an2n-1=120+(n-1)×1=n.∴an=n·2n-1,∴Sn=20+2×21+3×22+…+n×2n-1.①∴2Sn=21+2×22+…+(n-1)2n-1+n·2n.②∴①-②得-Sn=20+21+22+…+2n-1-n·2n=1-2n1-2-n·2n=(1-n)·2n-1.4∴Sn=(n-1)·2n+1.