1、能力升级练(八)数列求和与数列综合问题一、选择题1.已知数列an满足:任意m,nN*,都有anam=an+m,且a1=12,那么a5=()A.132B.116C.14D.12解析由题意,得a2=a1a1=14,a3=a1a2=18,则a5=a3a2=132.答案A2.(2023江西重点中学盟校联考)在数列an中,a1=-14,an=1-1an-1(n2,nN*),则a2 019的值为()A.-14B.5C.45D.54解析在数列an中,a1=-14,an=1-1an-1(n2,nN*),所以a2=1-1-14=5,a3=1-15=45,a4=1-145=-14,所以an是以3为周期的周期数列,
2、所以a2023=a6733=a3=45.答案C3.已知数列an的前n项和为Sn,且a1=2,an+1=Sn+1(nN*),则S5=()A.31B.42C.37D.47解析由题意,得Sn+1-Sn=Sn+1(nN*),Sn+1+1=2(Sn+1)(nN*),故数列Sn+1为等比数列,其首项为3,公比为2,则S5+1=324,所以S5=47.答案D4.(2023四川成都诊断)已知f(x)=(2a-1)x+4(x1),ax(x1),数列an(nN*)满足an=f(n),且an是递增数列,则a的取值范围是()A.(1,+)B.12,+C.(1,3)D.(3,+)解析因为an是递增数列,所以a1,a22
3、a-1+4,解得a3,则a的取值范围是(3,+).答案D5.(2017全国,理9)等差数列an的首项为1,公差不为0.若a2,a3,a6成等比数列,则an前6项的和为()A.-24B.-3C.3D.8解析设等差数列的公差为d,则d0,a32=a2a6,即(1+2d)2=(1+d)(1+5d),解得d=-2,所以S6=61+652(-2)=-24,故选A.答案A6.数列an的通项公式为an=(-1)n-1(4n-3),则它的前100项之和S100等于()A.200B.-200C.400D.-400解析S100=(41-3)-(42-3)+(43-3)-(4100-3)=4(1-2)+(3-4)+
4、(99-100)=4(-50)=-200.答案B7.数列an的通项公式是an=1n+n+1,前n项和为9,则n等于()A.9B.99C.10D.100解析因为an=1n+n+1=n+1-n,所以Sn=a1+a2+an=(n+1-n)+(n-n-1)+(3-2)+(2-1)=n+1-1,令n+1-1=9,得n=99.答案B8.(2023山东德州调研)已知Tn为数列2n+12n的前n项和,若mT10+1 013恒成立,则整数m的最小值为()A.1 026B.1 025C.1 024D.1 023解析2n+12n=1+12n,Tn=n+1-12n,T10+1013=11-1210+1013=1024
5、-1210,又mT10+1013恒成立,整数m的最小值为1024.答案C9.(2023福建厦门质检)已知数列an满足an+1+(-1)n+1an=2,则其前100项的和为()A.250B.200C.150D.100解析当n=2k(kN*)时,a2k+1-a2k=2,当n=2k-1(kN*)时,a2k+a2k-1=2,当n=2k+1(kN*)时,a2k+2+a2k+1=2,a2k+1+a2k-1=4,a2k+2+a2k=0,an的前100项和=(a1+a3)+(a97+a99)+(a2+a4)+(a98+a100)=254+250=100.答案D二、填空题10.若数列an的前n项和Sn=3n2-
6、2n+1,则数列an的通项公式an=.解析当n=1时,a1=S1=312-21+1=2;当n2时,an=Sn-Sn-1=3n2-2n+1-3(n-1)2-2(n-1)+1=6n-5,显然当n=1时,不满足上式.故数列的通项公式为an=2,n=1,6n-5,n2.答案2,n=1,6n-5,n211.在数列an中,a1=2,an+1n+1=ann+ln1+1n,则an=.解析由题意,得an+1n+1-ann=ln(n+1)-lnn,ann-an-1n-1=lnn-ln(n-1)(n2).a22-a11=ln2-ln1,a33-a22=ln3-ln2,ann-an-1n-1=lnn-ln(n-1)(
7、n2).累加,得ann-a11=lnn,ann=2+lnn(n2),又a1=2适合ann=2+lnn,故an=2n+nlnn.答案2n+nln n12.(2023湖北武汉质检)设数列(n2+n)an是等比数列,且a1=16,a2=154,则数列3nan的前15项和为.解析等比数列(n2+n)an的首项为2a1=13,第二项为6a2=19,故公比为13,所以(n2+n)an=1313n-1=13n,所以an=13n(n2+n),则3nan=1n2+n=1n-1n+1,其前n项和为1-1n+1,当n=15时,为1-116=1516.答案151613.等差数列an的前n项和记为Sn,若S44,S72
8、8,则a10的最大值为.解析等差数列an的前n项和为Sn,S44,S728,S4=4a1+432d4,S7=7a1+762d28,即2a1+3d2,a1+3d4,a10=a1+9d=a1+3d+6d4+6d,a10=a1+9d=12(2a1+3d)+15d22+15d2,2+15d2a104+6d,2+15d24+6d,解得d2,a104+62=16.答案16三、解答题14.求和Sn=x+1x2+x2+1x22+xn+1xn2(x0).解当x1时,Sn=x+1x2+x2+1x22+xn+1xn2=x2+2+1x2+x4+2+1x4+x2n+2+1x2n=(x2+x4+x2n)+2n+1x2+1
9、x4+1x2n=x2(x2n-1)x2-1+x-2(1-x-2n)1-x-2+2n=(x2n-1)(x2n+2+1)x2n(x2-1)+2n.当x=1时,Sn=4n.15.设数列an的前n项和为Sn,a1=2,an+1=2+Sn(nN*).(1)求数列an的通项公式;(2)设bn=1+log2(an)2,求证:数列1bnbn+1的前n项和Tn16.(1)解因为an+1=2+Sn(nN*),所以an=2+Sn-1(n2),所以an+1-an=Sn-Sn-1=an,所以an+1=2an(n2).又因为a2=2+a1=4,a1=2,所以a2=2a1,所以数列an是以2为首项,2为公比的等比数列,则an=22n-1=2n(nN*).(2)证明因bn=1+log2(an)2,则bn=2n+1.则1bnbn+1=1212n+1-12n+3,所以Tn=1213-15+15-17+12n+1-12n+3=1213-12n+3=16-12(2n+3)16.7
