考点过关检测(九)1.(2019·济宁模拟)已知数列{an}满足an+1=an-an-1(n≥2),a1=m,a2=n,Sn为数列{an}的前n项和,则S2019的值为()A.2019n-mB.n-2019mC.2mD.2n解析:选D∵an+1=an-an-1(n≥2),a1=m,a2=n,∴a3=n-m,a4=-m,a5=-n,a6=m-n,a7=m,a8=n,…,∴an+6=an,且a1+a2+a3+a4+a5+a6=0,则S2019=S336×6+3=336×(a1+a2+…+a6)+a1+a2+a3=336×0+m+n+n-m=2n.2.(2019·安徽马鞍山一模)已知函数f(n)=n2cos(nπ),且an=f(n)+f(n+1),则a1+a2+a3+…+a100=()A.0B.-100C.100D.10200解析:选Bf(n)=n2cos(nπ)==(-1)n·n2.由an=f(n)+f(n+1)=(-1)n·n2+(-1)n+1·(n+1)2=(-1)n[n2-(n+1)2]=(-1)n+1·(2n+1),得a1+a2+a3+…+a100=3+(-5)+7+(-9)+…+199+(-201)=-2×50=-100.故选B.3.(2019·泉州模拟)若数列{an}是正项数列,且++…+=n2+n,则a1++…+等于()A.2n2+2nB.n2+2nC.2n2+nD.2(n2+2n)解析:选A∵++…+=n2+n,∴n=1时,=2,解得a1=4.n≥2时,++…+=(n-1)2+n-1,相减可得=2n,∴an=4n2,n=1时也成立,∴=4n.则a1++…+=4(1+2+…+n)=4×=2n2+2n.4.(2019·广州模拟)已知递增数列{an}对任意n∈N*均满足an∈N*,aan=3n,记bn=a2·3n-1(n∈N*),则数列{bn}的前n项和等于()A.2n+nB.2n+1-1C.D.解析:选Daa1=3⇒a1≤3,讨论:若a1=1⇒aa1=a1=1,不合题意;若a1=2⇒a2=3;若a1=3⇒aa1=a3=3,不合题意,即a1=2,a2=3,aa2=6⇒a3=6,所以aa3=9⇒a6=9,所以a9=aa6=18,a18=aa9=27,a27=aa18=54,a54=aa27=81,则bn=3n,所以数列{bn}的前n项和等于=.5.(2019·河南郑州质检)已知数列{an}满足a1a2a3…an=2n2(n∈N*),且对任意n∈N*都有++…+的最大正整数n为________.解析:设等差数列{an}的公差为d,由已知可得解得故数列{an}的通项公式为an=2-n.Sn=a1++…+,①=++…+.②①-②得=a1++…+-=1--=1--=,所以Sn=,由Sn=>,得0