2019-2020学年数学北师大版必修5检测：1.2.2.2 an与Sn的关系及裂项求和法 .docx

第2课时an与Sn的关系及裂项求和法课后篇巩固探究A组1.已知数列an的前n项和Sn=1n,则a5的值等于()A.120B.-120C.130D.-130解析:a5=S5-S4=15-14=-120.答案:B2.已知等差数列an的前n项和为Sn,a5=5,S5=15,则数列1anan+1的前100项和为()A.100101B.99101C.99100D.101100解析:S5=5(a1+a5)2=5(a1+5)2=15,a1=1,d=a5-a15-1=5-15-1=1,an=1+(n-1)1=n,1anan+1=1n(n+1).设1anan+1的前n项和为Tn,则T100=112+123+1100101=1-12+12-13+1100-1101=1-1101=100101.答案:A3.设an(nN+)是等差数列,Sn是其前n项和,且S5<S6,S6=S7>S8,则下列结论错误的是()A.d<0B.a7=0C.S9>S5D.S6和S7均为Sn的最大值解析:由S5<S6得a1+a2+a5<a1+a2+a5+a6,a6>0.又S6=S7,a1+a2+a6=a1+a2+a6+a7,a7=0,故B正确;同理由S7>S8,得a8<0,又d=a7-a6<0,故A正确;由C选项中S9>S5,即a6+a7+a8+a9>0,可得2(a7+a8)>0.而由a7=0,a8<0,知2(a7+a8)>0不可能成立,故C错误;S5<S6,S6=S7>S8,S6与S7均为Sn的最大值,故D正确.故选C.答案:C4.数列1(n+1)2-1的前n项和Sn为()A.n+12(n+2)B.34-n+12(n+2)C.34-121n+1+1n+2D.32-1n+1-1n+2解析:1(n+1)2-1=1n2+2n=121n-1n+2,于是Sn=121-13+12-14+13-15+1n-1n+2=34-121n+1+1n+2.答案:C5.设函数f(x)满足f(n+1)=2f(n)+n2(nN+),且f(1)=2,则f(20)为()A.95B.97C.105D.192解析:f(n+1)=f(n)+n2,f(n+1)-f(n)=n2.f(2)-f(1)=12,f(3)-f(2)=22,f(20)-f(19)=192,f(20)-f(1)=1+2+192=(1+19)1922=95.又f(1)=2,f(20)=97.答案:B6.已知数列an的前n项和Sn=n2-9n,第k项满足5<ak<8,则k=.解析:an=Sn-Sn-1=(n2-9n)-(n-1)2-9(n-1)=2n-10(n2),又a1=S1=-8符合上式,所以an=2n-10.令5<2k-10<8,解得152<k<9.又kN+,所以k=8.答案:87.设数列an的前n项和为Sn,Sn=a1(3n-1)2,且a4=54,则a1=.解析:因为a4=S4-S3=a1(34-1)2-a1(33-1)2=27a1,所以27a1=54,解得a1=2.答案:28.数列1,11+2,11+2+3,11+2+3+n,的前n项和Sn=.解析:因为11+2+3+n=1n(n+1)2=2n(n+1)=21n-1n+1,所以Sn=1+11+2+11+2+3+11+2+3+n=21-12+12-13+13-14+1n-1n+1=21-1n+1=2nn+1.答案:2nn+19.正项数列an满足an2-(2n-1)an-2n=0.(1)求数列an的通项公式an;(2)令bn=1(n+1)an,求数列bn的前n项和Tn.解(1)由an2-(2n-1)an-2n=0,得(an-2n)(an+1)=0,即an=2n或an=-1,由于an是正项数列,故an=2n.(2)由(1)知an=2n,所以bn=1(n+1)an=12n(n+1)=121n-1n+1,故Tn=121-12+12-13+1n-1n+1=121-1n+1=n2(n+1).10.导学号33194014已知等差数列an的前n项和为Sn,nN+,且a3+a6=4,S5=-5.(1)求an;(2)若Tn=|a1|+|a2|+|a3|+|an|,求T5的值和Tn的表达式.解(1)设an的首项为a1,公差为d,易由a3+a6=4,S5=-5得出a1=-5,d=2.an=2n-7.(2)当n4时,an=2n-7>0;当n3时,an=2n-7<0,T5=-(a1+a2+a3)+a4+a5=13.当1n3时,Tn=-(a1+a2+an)=-n2+6n;当n4时,Tn=-(a1+a2+a3)+a4+a5+an=n2-6n+18.综上所述,Tn=-n2+6n,1n3,n2-6n+18,n4.B组1.若等差数列an的通项公式为an=2n+1,则由bn=a1+a2+ann所确定的数列bn的前n项之和是()A.n(n+2)B.12n(n+4)C.12n(n+5)D.12n(n+6)解析:由题意知a1+a2+an=n(3+2n+1)2=n(n+2),bn=n(n+2)n=n+2.于是数列bn的前n项和Sn=n(3+n+2)2=12n(n+5).答案:C2.已知一个等差数列共n项,且其前四项之和为21,末四项之和为67,前n项和为286,则项数n为()A.24B.26C.25D.28解析:设该等差数列为an,由题意,得a1+a2+a3+a4=21,an+an-1+an-2+an-3=67,又a1+an=a2+an-1=a3+an-2=a4+an-3,4(a1+an)=21+67=88,a1+an=22.Sn=n(a1+an)2=11n=286,n=26.答案:B3.已知数列an满足a1=1,an=an-1+2n(n2),则a7=()A.53B.54C.55D.109解析:an=an-1+2n,an-an-1=2n.a2-a1=4,a3-a2=6,a4-a3=8,an-an-1=2n(n2).an=1+4+6+2n=1+(n-1)(4+2n)2=n2+n-1.a7=72+7-1=55.答案:C4.已知数列an为12,13+23,14+24+34,110+210+310+910,如果bn=1anan+1,那么数列bn的前n项和Sn为()A.nn+1B.4nn+1C.3nn+1D.5nn+1解析:an=1+2+3+nn+1=n2,bn=1anan+1=4n(n+1)=41n-1n+1,Sn=41-12+12-13+1n-1-1n+1n-1n+1=41-1n+1=4nn+1.答案:B5.已知数列an的前n项和为Sn=n2+n+1,则an=.解析:当n=1时,a1=S1=3;当n2时,an=Sn-Sn-1=n2+n+1-(n-1)2+(n-1)+1=2n.此时,当n=1时,2n=23.所以an=3(n=1),2n(n2).答案:3(n=1),2n(n2)6.导学号33194015设Sn是等差数列an的前n项和,已知S6=36,Sn=324,若Sn-6=144(n>6),则数列的项数n为.解析:由题意可知a1+a2+a6=36,an+an-1+an-5=324-144,由+,得(a1+an)+(a2+an-1)+(a6+an-5)=216,6(a1+an)=216,a1+an=36.Sn=n(a1+an)2=18n=324,n=18.答案:187.设数列an的前n项和为Sn,a1=1,an=Snn+2(n-1)(nN+).(1)求证:数列an为等差数列,并求an与Sn;(2)是否存在自然数n,使得S1+S22+S33+Snn-(n-1)2=2 019?若存在,求出n的值;若不存在,请说明理由.(1)证明由an=Snn+2(n-1),得Sn=nan-2n(n-1)(nN+).当n2时,an=Sn-Sn-1=nan-(n-1)an-1-4(n-1),即an-an-1=4,故数列an是以1为首项,4为公差的等差数列.于是,an=4n-3,Sn=(a1+an)n2=2n2-n.(2)解存在自然数n使得S1+S22+S33+Snn-(n-1)2=2 019成立.理由如下:由(1),得Snn=2n-1(nN+),所以S1+S22+S33+Snn-(n-1)2=1+3+5+7+(2n-1)-(n-1)2=n2-(n-1)2=2n-1.令2n-1=2 019,得n=1 010,所以存在满足条件的自然数n为1 010.8.导学号33194016数列an的前n项和Sn=100n-n2(nN+).(1)求证an是等差数列;(2)设bn=|an|,求数列bn的前n项和.(1)证明an=Sn-Sn-1=(100n-n2)-100(n-1)-(n-1)2=101-2n(n2).a1=S1=1001-12=99=101-21,数列an的通项公式为an=101-2n(nN+).又an+1-an=-2为常数,数列an是首项a1=99,公差d=-2的等差数列.(2)解令an=101-2n0,得n50.5.nN+,n50(nN+).当1n50时an>0,此时bn=|an|=an,bn的前n项和Sn=100n-n2;当n51时an<0,此时bn=|an|=-an,由b51+b52+bn=-(a51+a52+an)=-(Sn-S50)=S50-Sn,得数列bn的前n项和为Sn=S50+(S50-Sn)=2S50-Sn=22 500-(100n-n2)=5 000-100n+n2.由得数列bn的前n项和为Sn=100n-n2(1n50,且nN+),5 000-100n+n2(n51,且nN+).