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專題對(duì)點(diǎn)練13 等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和
1.Sn為數(shù)列{an}的前n項(xiàng)和.已知an>0,an2+2an=4Sn+3.
(1)求{an}的通項(xiàng)公式;
(2)設(shè)bn=1anan+1,數(shù)列{bn}的前n項(xiàng)和為Tn,求Tn.
解 (1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.兩式相減可得an+12-an2+2(an+1-an)=4an+1,即2(an+1+an)=an+12-an2=(an+1+an)(an+1-an).由于an>0,因此an+1-an=2.
又a12+2a1=4a1+3,解得a1=3(a1=-1舍去).
所以{a
2、n}是首項(xiàng)為3,公差為2的等差數(shù)列,故an=2n+1.
(2)由an=2n+1可知bn=1anan+1=1212n+1-12n+3.
Tn=b1+b2+…+bn=1213-15+15-17+…+12n+1-12n+3=n3(2n+3).
2.已知數(shù)列{an}是等差數(shù)列,前n項(xiàng)和為Sn,若a1=9,S3=21.
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)若a5,a8,Sk成等比數(shù)列,求k的值.
解 (1)設(shè)等差數(shù)列{an}的公差為d,∵a1=9,S3=21,
∴S3=3×9+3×22d=21,解得d=-2,
∴an=9+(n-1)×(-2)=-2n+1
3、1.
(2)∵a5,a8,Sk成等比數(shù)列,∴a82=a5·Sk,即(-2×8+11)2=(-2×5+11)·9k+k(k-1)2×(-2),解得k=5.
3.(2017河北衡水中學(xué)三調(diào),理17)已知數(shù)列{an}的前n項(xiàng)和為Sn,a1≠0,常數(shù)λ>0,且λa1an=S1+Sn對(duì)一切正整數(shù)n都成立.
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)設(shè)a1>0,λ=100,當(dāng)n為何值時(shí),數(shù)列l(wèi)g1an的前n項(xiàng)和最大?
解 (1)令n=1,得λa12=2S1=2a1,即a1(λa1-2)=0.
因?yàn)閍1≠0,所以a1=2λ.
當(dāng)n≥2
4、時(shí),2an=2λ+Sn,2an-1=2λ+Sn-1,
兩式相減,得2an-2an-1=an(n≥2),
所以an=2an-1(n≥2),從而數(shù)列{an}為等比數(shù)列,
所以an=a1·2n-1=2nλ.
(2)當(dāng)a1>0,λ=100時(shí),由(1)知,an=2n100,
設(shè)bn=lg1an=lg1002n=lg 100-lg 2n=2-nlg 2,
所以數(shù)列{bn}是單調(diào)遞減的等差數(shù)列,公差為-lg 2,
所以b1>b2>…>b6=lg10026=lg10064>lg 1=0,
當(dāng)n≥7時(shí),bn≤b7=lg10027<lg 1=0,
所
5、以數(shù)列l(wèi)g1an的前6項(xiàng)和最大.
4.(2017河北邯鄲二模,理17)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a1≠0,a3=3,且λSn=anan+1.在等比數(shù)列{bn}中,b1=2λ,b3=a15+1.
(1)求數(shù)列{an}及{bn}的通項(xiàng)公式;
(2)設(shè)數(shù)列{cn}的前n項(xiàng)和為Tn,且Sn+n2cn=1,求Tn.
解 (1)∵λSn=anan+1,a3=3,∴λa1=a1a2,且λ(a1+a2)=a2a3,
∴a2=λ,a1+a2=a3=3.①
∵數(shù)列{an}是等差數(shù)列,∴a1+a3=2a2,即2a2-a1=3.②
由①②得a1=1,a2=2,∴an=n,λ=2,∴b1=4,
6、b3=16,
∴{bn}的公比q=±b3b1=±2,
∴bn=2n+1或bn=(-2)n+1.
(2)由(1)知Sn=n(1+n)2,∴cn=2n(n+2)=1n-1n+2,
∴Tn=1-13+12-14+13-15+…+1n-1-1n+1+1n-1n+2=1+12-1n+1-1n+2=32-2n+3n2+3n+2.
5.(2017寧夏中衛(wèi)二模,理17)已知等比數(shù)列{an}的公比q>1,且a1+a3=20,a2=8.
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)設(shè)bn=nan,Sn是數(shù)列{bn}的前n項(xiàng)和,求Sn.
解 (1)∵等比數(shù)列{an}的公比q&g
7、t;1,且a1+a3=20,a2=8,
∴a1+a1q2=20,a1q=8,∴2q2-5q+2=0,解得q=2,a1=4.∴an=2n+1.
(2)bn=nan=n2n+1,Sn=122+223+…+n2n+1,
12Sn=123+224+…+n-12n+1+n2n+2.
∴12Sn=122+123+…+12n+1-n2n+2=141-12n1-12-n2n+2=12-2+n2n+2.∴Sn=1-2+n2n+1.
6.(2017安徽安慶二模,理17)在數(shù)列{an}中,a1=2,a2=4,設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,對(duì)于任意的n>1,n∈N*,Sn+1+Sn-1=2(Sn+1
8、).
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)設(shè)bn=n2an,求{bn}的前n項(xiàng)和Tn.
解 (1)對(duì)于任意的n>1,n∈N*,Sn+1+Sn-1=2(Sn+1),Sn+2+Sn=2(Sn+1+1),
兩式相減可得an+2+an=2an+1.(*)
當(dāng)n=2時(shí),S3+S1=2(S2+1),即2a1+a2+a3=2(a1+a2+1),解得a3=6.∴當(dāng)n=1時(shí)(*)也滿足.
∴數(shù)列{an}是等差數(shù)列,公差為2,
∴an=2+2(n-1)=2n.
(2)∵bn=n2an=n22n=n4n,∴Tn=14+242+343+…+n4n,14Tn=142+243+…+n-14n+n
9、4n+1,
∴34Tn=14+142+…+14n-n4n+1=141-14n1-14-n4n+1,
∴Tn=49-4+3n9×4n. ?導(dǎo)學(xué)號(hào)16804189?
7.(2017山東,理19)已知{xn}是各項(xiàng)均為正數(shù)的等比數(shù)列,且x1+x2=3,x3-x2=2.
(1)求數(shù)列{xn}的通項(xiàng)公式;
(2)如圖,在平面直角坐標(biāo)系xOy中,依次連接點(diǎn)P1(x1,1),P2(x2,2)…Pn+1(xn+1,n+1)得到折線P1P2…Pn+1,求由該折線與直線y=0,x=x1,x=xn+1所圍成的區(qū)域的面積Tn.
解 (1)設(shè)數(shù)列{xn}的公比為q,由已知q>0.
由題
10、意得x1+x1q=3,x1q2-x1q=2.所以3q2-5q-2=0.
因?yàn)閝>0,所以q=2,x1=1,
因此數(shù)列{xn}的通項(xiàng)公式為xn=2n-1.
(2)過(guò)P1,P2,…,Pn+1向x軸作垂線,垂足分別為Q1,Q2,…,Qn+1.
由(1)得xn+1-xn=2n-2n-1=2n-1,
記梯形PnPn+1Qn+1Qn的面積為bn,
由題意bn=(n+n+1)2×2n-1=(2n+1)×2n-2,
所以Tn=b1+b2+…+bn=3×2-1+5×20+7×21+…+(2n-1)×2n-3+(2n+1)×
11、2n-2. ①
又2Tn=3×20+5×21+7×22+…+(2n-1)×2n-2+(2n+1)×2n-1, ②
①-②得-Tn=3×2-1+(2+22+…+2n-1)-(2n+1)×2n-1=32+2(1-2n-1)1-2-(2n+1)×2n-1.
所以Tn=(2n-1)×2n+12. ?導(dǎo)學(xué)號(hào)16804190?
8.(2017山東濰坊一模,理19)已知數(shù)列{an}是等差數(shù)列,其前n項(xiàng)和為Sn,數(shù)列{bn}是公比大于0的等比數(shù)列,且b1=-2a1=2,a3+b2=-1,S3+2b3=7.
(1
12、)求數(shù)列{an}和{bn}的通項(xiàng)公式;
(2)令cn=2,n為奇數(shù),-2anbn,n為偶數(shù),求數(shù)列{cn}的前n項(xiàng)和Tn.
解 (1)設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q,q>0,
∵b1=-2a1=2,a3+b2=-1,S3+2b3=7,
∴a1=-1,b1=2,-1+2d+2q=-1,3×(-1)+3d+2×2×q2=7,
解得d=-2,q=2.∴an=-1-2(n-1)=1-2n,bn=2n.
(2)cn=2,n為奇數(shù),2n-12n-1,n為偶數(shù).
①當(dāng)n=2k(k∈N*)時(shí),數(shù)列{cn}的前n項(xiàng)和Tn=T2k=(c1
13、+c3+…+c2k-1)+(c2+c4+…+c2k)
=2k+32+723+…+4k-122k-1,
令A(yù)k=32+723+…+4k-122k-1,
∴14Ak=323+725+…+4k-522k-1+4k-122k+1,
∴34Ak=32+4123+125+…+122k-1-4k-122k+1=32+4×181-14k-11-14-4k-122k+1,
∴Ak=269-12k+139×22k-1.∴Tn=T2k=2k+269-12k+139×22k-1.
②當(dāng)n=2k-1(k∈N*)時(shí),數(shù)列{cn}的前n項(xiàng)和Tn=T2k-2+a2k-1=2(k-1)+269-12(k-1)+139×22(k-1)-1+2=2k+269-12k+19×22k-3.
∴Tn=2k+269-12k+139×22k-1,n=2k,2k+269-12k+19×22k-3,n=2k-1,k∈N*.
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高考數(shù)學(xué)二輪復(fù)習(xí) 專題對(duì)點(diǎn)練13 等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和 理
