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,n∈N*,都有an·am=an+m,且a1=12,那么a5=()A.132B.116C.14D.12解析由题意,得a2=a1a1=14,a3=a1·a2=18,则a5=a3·a2=132.答案A2.(2019江西重点中学盟校联考)在数列{an}中,a1=-14,an=1-1an-1(n≥2,n∈N*),则a2019的值为()A.-14B.5C.45D.54解析在数列{an}中,a1=-14,an=1-1an-1(n≥2,n∈N*),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,所以a2019=a673×3=a3=45.答案C3.已知数列{an}的前n项和为Sn,且a1=2,an+1=Sn+1(n∈N*),则S5=()A.31B.42C.37D.47解析由题意,得Sn+1-Sn=Sn+1(n∈N*),∴Sn+1+1=2(Sn+1)(n∈N*),故数列{Sn+1}为等比数列,其首项为3,公比为2,则S5+1=3×24,所以S5=47.答案D4.(2019四川成都诊断)已知f(x)={(2a-1)x+4(x≤1),ax(x>1),数列{an}(n∈N*)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{a>1,a2>2a-1+4,解得a>3,则a的取值范围是(3,+∞).答案D5.(2017全国Ⅲ,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d≠0,a32=a2·a6,即(1+2d)2=(1+d)(1+5d),解得d=-2,所以S6=6×1+6×52×(-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=(4×1-3)-(4×2-3)+(4×3-3)-…-(4×100-3)=4×[(1-2)+(3-4)+…+(99-100)]=4×(-50)=-200.答案B7.数列{an}的通项公式是an=1❑√n+❑√n+1,前n项和为9,则n等于()A.9B.99C.10D.100解析因为an=1❑√n+❑√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.(2019山东德州调研)已知Tn为数列{2n+12n}的前n项和,若m>T10+1013恒成立,则整数m的最小值为()A.1026B.1025C.1024D.1023解析∵2n+12n=1+(12)n,∴Tn=n+1-12n,∴T10+1013=11-1210+1013=1024-1210,又m>T10+1013恒成立,∴整数m的最小值为1024.答案C9.(2019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(k∈N*)时,a2k+1-a2k=2,当n=2k-1(k∈N*)时,a2k+a2k-1=2,当n=2k+1(k∈N*)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)=25×4+25×0=100.答案D二、填空题10.若数列{an}的前n项和Sn=3n2-2n+1,则数列{an}的通项公式an=.解析当n=1时,a1=S1=3×12-2×1+1=2;当n≥2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,n≥2.答案{2,n=1,6n-5,n≥211.在数列{an}中,a1=2,an+1n+1=ann+ln(1+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)(n≥2).∴a22−a11=ln2-ln1,a33−a22=ln3-ln2,…,ann−an-1n-1=lnn-ln(n-1)(n≥2).累加,得ann−a11=lnn,∴ann=2+lnn(n≥2),又a1=2适合ann=2+lnn,故an=2n+nlnn.答案2n+nlnn12.(2019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=13·(13)n-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S4≥4,S7≤28,则a10的最大值为.解析∵等差数列{an}的前n项和为Sn,S4≥4,S7≤28,∴{S4=4a1+4×32d≥4,S7=7a1+7×62d≤28,即{2a1+3d≥2,a1+3d≤4,∴{a10=a1+9d=a1+3d+6d≤4+6d,a10=a1+9d=12(2a1+3d)+15d2≥2+15d2,∴2+15d2≤a10≤4+6d,∴2+15d2≤4+6d,解得d≤2,∴a10≤4+6×2=16.答案16三、解答题14.求和Sn=(x+1x)2+(x2+1x2)2+…+(xn+1xn)2(x≠0).解当x≠±1时,Sn=(x+1x)2+(x2+1x2)2+…+(xn+1xn)2=x2+2+1x2+x4+2+1x4+…+x2n+2+1x2n=(x2+x4+…+x2n)+2n+1x2+1x4+…+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(n∈N*).(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Tn<16.(1)解因为an+1=2+Sn(n∈N*),所以an=2+Sn-1(n≥2),所以an+1-an=Sn-Sn-1=an,所以an+1=2an(n≥2).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=2·2n-1=2n(n∈N*).(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.
