2021新高考数学二轮总复习专题突破练15　求数列的通项及前N项和 WORD版含解析.docx

1、专题突破练15求数列的通项及前n项和1.已知等差数列an的前n项和为Sn,且a1=1,S3+S4=S5.(1)求数列an的通项公式;(2)令bn=(-1)n-1an,求数列bn的前n项和Tn.2.(2020山东滨州二模,18)已知an为等差数列,a3+a6=25,a8=23,bn为等比数列,且a1=2b1,b2b5=a11.(1)求an,bn的通项公式;(2)记cn=anbn,求数列cn的前n项和Tn.3.(2020全国,理17)设数列an满足a1=3,an+1=3an-4n.(1)计算a2,a3,猜想an的通项公式并加以证明;(2)求数列2nan的前n项和Sn.4.(2020山东聊城二模,1

2、7)已知数列an的各项均为正数,其前n项和为Sn,且an2+an=2Sn+34(nN*).(1)求数列an的通项公式;(2)若bn=1Sn,求bn的前n项和Tn.5.(2020山东青岛5月模拟,17)设数列an的前n项和为Sn,a1=1,给出下列三个条件;条件:数列an为等比数列,数列Sn+a1也为等比数列;条件:点(Sn,an+1)在直线y=x+1上;条件:2na1+2n-1a2+2an=nan+1.试在上面的三个条件中任选一个,补充在上面的横线上,完成下列两问的解答:(1)求an的通项公式;(2)设bn=1log2an+1log2an+3,求数列bn的前n项和Tn.6.(2020山东菏泽一

3、模,18)已知数列an满足nan+1-(n+1)an=1(nN*),且a1=1.(1)求数列an的通项公式;(2)若数列bn满足bn=an3n-1,求数列bn的前n项和Sn.7.已知数列an的前n项和Sn=3n2+8n,bn是等差数列,且an=bn+bn+1.(1)求数列bn的通项公式;(2)令cn=(an+1)n+1(bn+2)n,求数列cn的前n项和Tn.8.(2020天津南开区一模,18)已知数列an的前n项和Sn=n2+n2,数列bn满足:b1=b2=2,bn+1bn=2n+1(nN*).(1)求数列an,bn的通项公式;(2)求i=1naib2i-1-1b2i(nN*).专题突破练1

4、5求数列的通项及前n项和1.解(1)设等差数列an的公差为d,由S3+S4=S5可得a1+a2+a3=a5,即3a2=a5,3(1+d)=1+4d,解得d=2.an=1+(n-1)2=2n-1.(2)由(1)可得bn=(-1)n-1(2n-1).T2n=1-3+5-7+(2n-3)-(2n-1)=(-2)n=-2n.当n为偶数时,Tn=-n;当n为奇数时,Tn=Tn-1+bn=-(n-1)+(-1)n-1an=-(n-1)+(-1)n-1(2n-1)=-(n-1)+(2n-1)=n.综上,Tn=(-1)n+1n.2.解(1)设等差数列an的公差为d,由题意得2a1+7d=25,a1+7d=23

5、,解得a1=2,d=3.所以数列an的通项公式an=3n-1.设等比数列bn的公比为q,由a1=2b1,b2b5=a11,得b1=1,b12q5=32,解得q=2,所以数列bn的通项公式bn=2n-1.(2)由(1)知,cn=anbn=(3n-1)2n-1,则Tn=c1+c2+c3+cn-1+cn=220+521+822+(3n-4)2n-2+(3n-1)2n-1,2Tn=221+522+823+(3n-4)2n-1+(3n-1)2n.两式相减得-Tn=2+3(21+22+2n-1)-(3n-1)2n=2+32-2n-121-2-(3n-1)2n=-4+(4-3n)2n,所以Tn=4+(3n-

6、4)2n.3.解(1)a2=5,a3=7.猜想an=2n+1.由已知可得an+1-(2n+3)=3an-(2n+1),an-(2n+1)=3an-1-(2n-1),a2-5=3(a1-3).因为a1=3,所以an=2n+1.(2)由(1)得2nan=(2n+1)2n,所以Sn=32+522+723+(2n+1)2n.从而2Sn=322+523+724+(2n+1)2n+1.-得-Sn=32+222+223+22n-(2n+1)2n+1.所以Sn=(2n-1)2n+1+2.4.解(1)因为an2+an=2Sn+34(nN*),所以当n2时,an-12+an-1=2Sn-1+34,-得an2-an

7、-12+an-an-1=2(Sn-Sn-1),即an2-an-12-an-an-1=0,所以(an+an-1)(an-an-1-1)=0.因为an0,所以an-an-1=1,所以数列an是公差为1的等差数列,当n=1时,由a12+a1=2S1+34可得,a1=32,所以an=a1+(n-1)d=32+(n-1)1=n+12.(2)由(1)知Sn=na1+n(n-1)2d=n(n+2)2,所以bn=1Sn=2n(n+2)=1n-1n+2,Tn=b1+b2+b3+bn-1+bn=11-13+12-14+13-15+1n-1-1n+1+1n-1n+2=32-1n+1-1n+2=3n2+5n2n2+6

8、n+4.5.解(1)方案一:选条件.因为数列Sn+a1为等比数列,所以(S2+a1)2=(S1+a1)(S3+a1),即(2a1+a2)2=2a1(2a1+a2+a3).设等比数列an的公比为q,因为a1=1,所以(2+q)2=2(2+q+q2),解得q=2或q=0(舍).所以an=a1qn-1=2n-1(nN*).方案二:选条件.因为点(Sn,an+1)在直线y=x+1上,所以an+1=Sn+1(nN*),所以an=Sn-1+1(n2).两式相减得an+1-an=an,an+1an=2(n2).因为a1=1,a2=S1+1=a1+1=2,a2a1=2适合上式,所以数列an是首项为1,公比为2

9、的等比数列,所以an=a1qn-1=2n-1(nN*).方案三:选条件.当n2时,因为2na1+2n-1a2+2an=nan+1(nN*),所以2n-1a1+2n-2a2+2an-1=(n-1)an.所以2na1+2n-1a2+22an-1=2(n-1)an.-得2an=nan+1-2(n-1)an,即an+1an=2(n2).当n=1时,2a1=a2,a2a1=2适合上式,所以数列an是首项为1,公比为2的等比数列,所以an=a1qn-1=2n-1(nN*).(2)由(1)得an=2n-1(nN*),所以bn=1log2an+1log2an+3=1n(n+2)=121n-1n+2.所以Tn=

10、121-13+12-14+13-15+1n-1-1n+1+1n-1n+2=1232-1n+1-1n+2=34-121n+1+1n+2=34-2n+32(n+1)(n+2).6.解(1)因为nan+1-(n+1)an=1,所以an+1n+1-ann=1n(n+1)=1n-1n+1,所以ann-an-1n-1=1n-1-1n(n2),an-1n-1-an-2n-2=1n-2-1n-1,a22-a11=1-12,所以ann-a1=1-1n(n2).又a1=1,所以ann=2n-1n,所以an=2n-1(n2).又a1=1,也符合上式,所以对任意正整数n,an=2n-1.(2)结合(1)得bn=2n-

11、13n-1,所以Sn=130+331+532+733+2n-13n-1,13Sn=13+332+533+2n-13n,-,得23Sn=1+213+132+13n-1-2n-13n=2-2n+23n,所以Sn=3-n+13n-1.7.解(1)由题意知当n2时,an=Sn-Sn-1=6n+5,当n=1时,a1=S1=11,所以an=6n+5.设数列bn的公差为d,由a1=b1+b2,a2=b2+b3,即11=2b1+d,17=2b1+3d,可解得b1=4,d=3,所以bn=3n+1.(2)由(1)知cn=(6n+6)n+1(3n+3)n=3(n+1)2n+1,又Tn=c1+c2+c3+cn,得Tn

12、=3222+323+424+(n+1)2n+1,2Tn=3223+324+425+(n+1)2n+2,两式作差,得-Tn=3222+23+24+2n+1-(n+1)2n+2=34+4(2n-1)2-1-(n+1)2n+2=-3n2n+2,所以Tn=3n2n+2.8.解(1)当n2时,an=Sn-Sn-1=n2+n2-(n-1)2+(n-1)2=n,n=1时,a1=S1=1,满足上式,an=n.bn+1bn=2n+1,bnbn-1=2n(n2),bn+1=2bn-1(n2),数列bn的奇数项和偶数项分别是2为首项,2为公比的等比数列,bn=2n+12,n为奇数,2n2,n为偶数.(2)i=1naib2i-1-1b2i=i=1ni2i-12i=i=1ni2i-i=1ni2i,设Mn=1x+2x2+3x3+(n-1)xn-1+nxn(x0,1),xMn=1x2+2x3+3x4+(n-1)xn+nxn+1,-得(1-x)Mn=x+x2+x3+xn-nxn+1=x(1-xn)1-x-nxn+1,Mn=x+(nx-n-1)xn+1(1-x)2.i=1ni2i=2+(2n-n-1)2n+1(1-2)2=(n-1)2n+1+2,i=1ni2i=12+n2-n-112n+11-122=2-n+22n,从而i=1naib2i-1-1b2i=(n-1)2n+1+n+22n.