錯(cuò)位相減法(提高篇)

數(shù)列求和之錯(cuò)位相減法的1］已知數(shù)列｛an｝的前n項(xiàng)和為Sn,且有a1=2,3Sn=5an-4an_1+3Sn/n>2)(I)求數(shù)列an的通項(xiàng)公式；(II)若bn=nq,求數(shù)列｛bn｝的前n項(xiàng)和Tn。n n_1 n n_1 n n_1an_1又a=2，.乂a｝是以n n_1 n n_1 n n_1an_1又a=2，.乂a｝是以2為首項(xiàng)，2為公比的等比數(shù)列， (4分)1n二.a=2-2n_i=2n. (5 分)n(II)b=n-2n,T=1-2i+2-22+3-23+ +n-2n,TOC\o"1-5"\h\z2T=1-22+2-23++(n_1)-2n+n-2n+i (8 分)兩式相減得：—T=2i+22+ +2n_n?2n+i,2(1_2n).?.—T= n?2n+i=(i—n)?2n+i—2, (11 分)ni_2??.T=2+(n—i)?2n+i (12 分)［例2］等比數(shù)列包｝的前n項(xiàng)和為Sn.已知于S3,S2成等差數(shù)列.(1)求小)的公比q;3⑵若a［a3=-2，求數(shù)列｛n?an｝的刖n項(xiàng)和Tn.解析：(1)由已知得253二1+邑/-2(a1+a2+a3)=a1+(a1+a2),「.a2+2a3=0,anW0,1「.1+2q=0,「.q=_2.13 3(2)?「Wa1(1-q2)=a1”?二已二一],1 1??.4二-2,「.叫二(-2).(-2)nT=(-2)n-2,1二nan=n(-2)n-2T1 1 1 1AA??Tn=1(-2)-1+2(-2)o+3(-2)1+'''+n?(-2)n-2，①TOC\o"1-5"\h\z1丁 1 1 1 1 …」?一2Tn=1?(-2)o+2?(-2)1+3?(-2)2+…+n-(-2)n-1,②①-②得3 111 1 12Tn=-2+[(-2)0+(-2)1+(-2)2+…+(-2)n-2]-n(-2)n-14 12=-3-(-2)nT(3+n)，8 142?I二-9-(-2)n-1(9+3n).[例3]設(shè)數(shù)列｛an｝滿足a1=2,an+1-an=3-2^-1.⑴求數(shù)列｛an｝的通項(xiàng)公式；⑵令bn二nan,求數(shù)列｛bn｝的前n項(xiàng)和Sn.解析(1)由已知，得當(dāng)nN1時(shí)，an+1=[(an+1-an)+(丫4-)+…+(2-2)]+J二3(22n-1+22n-3+…+2)+2=22(n+1)T.而a「2,符合上式，所以數(shù)列｛an｝的通項(xiàng)公式為an=22n-1.

⑵由bn二nan=n2nT知Sn=1-2+223+325+…+n22n-i.①從而2 2 + 4+ +2n.?Sn=L23+2^25+3^27+ 2 + 4+ +2n.①—②得(1—22)S=2+23+25+…+22n-1—n.22n+i,1即Sn=9[(3n-1)22n+1+2].J[例4]已知等差數(shù)列｛an｝滿足a「0,a6+a8=-10.⑴求數(shù)列｛an｝的通項(xiàng)公式；a⑵求數(shù)列｛叱的前n項(xiàng)和.fa+d=0,解析(1)設(shè)等差數(shù)列｛an｝的公差為d.由已知條件可得［2a+12d二-10,fa=1,解得比-1故數(shù)列編的通項(xiàng)公式為an=2-Q⑵設(shè)數(shù)列12y的前n項(xiàng)和為Sn,即Sn=21+%,+畀,故S「LSa a af二-1 2 nS所以，當(dāng)n>1時(shí),了二a「a-a

—212+…+an—an1

2n-1a11 1子1-(2+4+3-2-n〒二11-q一不)2-nn n-Q二^".所以S=~~^.綜上，2n2n n2n-1,,az n數(shù)列124的前n項(xiàng)和Sn二不22a n,n-1,2,3，….a+1[例5](2008,XX)已知數(shù)歹ij｛an｝的首項(xiàng)a1=|,an+11(I)證明：數(shù)歹M--1)是等比數(shù)列；

a（II）數(shù)列｛一｝的前n項(xiàng)和San解析（工）匕+12a n-

a+11a+1111an+1an+111二數(shù)列｛——1］是以為個(gè)首項(xiàng)12為公比的等比數(shù)列.（口）由（工）an+111二數(shù)列｛——1］是以為個(gè)首項(xiàng)12為公比的等比數(shù)列.（口）由（工）11知一-1-萬112n-12n11即---+1a2nnn=—+n2n+ + +2 22 2312--+—+22 232n+1——+ +2221卜 一\o"CurrentDocument"2n 2n+11-1、2(r^3T22n+1-1-1—工2n 2n+111-1-5(T)2an n n、— —.數(shù)列｛—｝的前n項(xiàng)和

an［例6］在等比數(shù)列州中,a2a3=32,4二32,1n n(n+1)???T-2—————.又1+2+3+…+n-———-

n 2n-1 2n 22+n+n(n+1) n2+n+4n+2⑴求數(shù)列｛an｝的通項(xiàng)公式⑵設(shè)數(shù)列｛an｝的前n項(xiàng)和為Sn,求S1+2s2+…+nSn.解析：⑴設(shè)等比數(shù)列｛an｝的公比為q,依題意得faiq-aiq2=3211alq-32,解得21=2,q=2,故a=22-1=2n.n⑵表示數(shù)列同｝的前口項(xiàng)和,-2(1-2n)一?Sn二十一二二2(2nT)，「.S1+2S2+…+nSn=2[(2+22+…+n2)-(1+2+…+n)]=2(2+2-22+…+n-2n)-n(n+1)，設(shè)T「2+2?22+-+n-2n,?貝IJ2Tn=22+22+-+n2+i,②①-②，得2(1-2n)T=2+22+…+2n—n?2n+1——~~—n-2n+1—(1-n)2n+1—2,n 1-2?.Tn-(n-1)2n+1+2,「.S1+2S2+…+nSn―2[(n-1)2n+1+2]-n(n+1)(n-1)2n+2+4-n(n+1).[例7]已知各項(xiàng)均為正數(shù)的數(shù)列"J前n項(xiàng)和為S，首項(xiàng)為a且1,a,S等差數(shù)列.fn In tvtv(I)求數(shù)列LJ的通項(xiàng)公式；TOC\o"1-5"\h\z(II)若a2=(1)bn，設(shè)c=b，求數(shù)列4｝的前n項(xiàng)和T.n2 na n nn1 c 八解析：(1)由題意知2an=S+-,a>0 1分nn2n=2an-l當(dāng)〃22時(shí)S=2a—=2an-l=WJ'nn2“t兩式相減得T=S〃-Sj=2a，~兩式相減得T=S〃-Sj=2a，~2Cln-l2為公比的等比數(shù)歹5=a-2"—i=—x2"—1=2n-21 2整理得：=2“2=“2=2~bn=22〃-4nb—w-an4-2n_16-8n2n-2b—w-an4-2n_16-8n2n-227?—+2+士…n222 2324-8〃16-8〃 + 2〃T2〃1 8 01 8 0 24—8〃16—8〃—T=——+—+...+ + 2n22232"+i①-②得a="8(1+￡+...+/①-②得a="8(1+￡+...+/16-8n2"+1—(1--=4-8.2?一瀉1--—(1--=4-8.2?一瀉1--2)16-8n2"+1=4-4(1-—)-

2"-i16-Sn=4-4(1-—)-

2"-i16-Sn2?+i4〃4〃2n8n128n12分[例8](14分)已知單調(diào)遞增的等比數(shù)列｛a｝滿足q+aja「28,且a,+2是a2,的等差中n / 3 4 o NT項(xiàng).⑴求數(shù)列｛aj的通項(xiàng)公式；⑵若b二alog場,S=b+b+…+b,求使S+n,2n+i>50成立的最小正整數(shù)n的值.nnnn1 2 n n2解析⑴設(shè)此等比數(shù)列為ara^,a/2,a/3,…，其中aiW0,qWQ由題意知:a】q+a/2+a/3=28,①21+2/3=2年02+2).②②X7-①得6alq3—15alq2+6a7=0,1即2q2-5q+2=0,解得q=2或q=Q.???等比數(shù)列州單調(diào)遞增，.3=2,q=2,.*.an=2n.(2)由(1)得bn=-n?2%.*.S=b+b+ +b=-(1X2+2X22+???+n?2n).n1 2 n' /設(shè)T=1X2+2X22+…+n?2n,③n貝I」2T=IX22+2X23+…+n?2n+i.④n由③—④，得—'=1X2+1X22+…+1?2n—n?2n+l=2n+1-2-n,2n+1=(1-n)?2n+1-2,-T=一(n—1)?2n+i—2.nSn=-(n-1)?2n+i-2.要使S+n,2n+i>50成立，nBP-(n-1),2n+1-2+n,2n+1>50,即2n>26.,/24=16<26,25=32>26,且y=2x是單調(diào)遞增函數(shù),

???滿足條件的n的最小值為5.【跟蹤訓(xùn)練】1.已知數(shù)列｛an｝的前n項(xiàng)和為S「3n,數(shù)列｛bn｝滿足b「-1,bn+i=bn+(2n-1)(neN*).⑴求數(shù)列｛an｝的通項(xiàng)公式an;⑵求數(shù)列｛bn｝的通項(xiàng)公式bn;ab⑶若二V，求數(shù)列｛｝的前n項(xiàng)和Tn.解析：(1)?.?S=3n,「.S=3n-i(nN2),「.a二S-S =3口-3n-i=2X3n-i(nN2).n n-1 nnn-1當(dāng)n=1當(dāng)n=1時(shí)，2X31-1=2WS1=a1=3,「.an(3,n=1,12X3n-1,nN2.(2)：bn+1= bn+ (2nT),.「b2-b1= 1, b3-b2 =3, b4-b3=5,…，bn-bn 1= 2n-3.以上各式相加得bn以上各式相加得bn-b]=1+3+5+…+(2n-3)=(n-1)(1+2n-3)二(n-1)2.??七二-1，"二n2-2n.I-3,n=1,⑶由題意得二j〔2(n-2)X3n-1,n三2.當(dāng)nN2時(shí)，Tn=-3+2X0X31+2X1X32+2X2X33+…+2(n-2)X3n-1,「.3Tn=-9+2X0X32+2X1X33+2X2X34+…+2(n-2)X3n,「相減得-2Tn=6+2X32+2X33+…+2X3n-1-2(n-2)X3n.3n-3(2n-5)3n+3???Tn=(n-2)X3n-(3+32+33+?一+3n-1)=(n-2)X3n--2-= ~2 ?「Tn=?「Tn=](2n-5)3n+312nN2.「?Tn(2n-5)3n+3

2(nEN*).2.已知數(shù)列｛an｝為公差不為零的等差數(shù)列，？二1,各項(xiàng)均為正數(shù)的等比數(shù)列｛bn｝的第1項(xiàng),第3項(xiàng)，第5項(xiàng)分別是a1,a3,a2i.⑴求數(shù)列｛an｝與｛bn｝的通項(xiàng)公式；⑵求數(shù)列｛anbn｝的前n項(xiàng)和Sn.解析：⑴設(shè)數(shù)列｛an｝的公差為d(dWO),數(shù)列｛bn｝的公比為q,??由題意得23二a1a2i,/.(1+2d)2=1X(1+20d),即4d2—16d=0,?,dW0,「.d=4,「.an=4n-3.?.b「1,b「9,b「81,???｛bn｝的各項(xiàng)均為正數(shù)，「?q=3,「.bn=3n-1.(2)?.?由(1)可得anbn=(4n-3)3n-1,??.Sn=3o+5X31+9x32+…+(4n-7)X3n-2+(4n-3)X3n-1,3Sn=31+5X32+9X33+…+(4n-7)X3n-1+(4n-3)X3n,兩式相減得：-2Sn=1+4X3+4X32+4X33+…+4X3n-1-(4n-3)X3n=1+4(3+32+33+…+3n-1)-(4n-3)X3n1+4X3X(1+4X3X(1-3n-1)

1-3-(4n-3)X3n二(5-4n)X3n-5,??.S二??.S二n(4n-5)3n+53、已知遞增的等比數(shù)列｛an｝滿足：a2+a3+a4=28,且a3+2是a2,a4的等差中項(xiàng).⑴求數(shù)列｛an｝的通項(xiàng)公式;- 1⑵若以二斗崛2a「，3二bi+b2+-+bn，求S’解：⑴設(shè)等比數(shù)列｛an｝的首項(xiàng)為q，公比為q.依題意，有2(a3+2)=a2+a4,代入22+23+2廠28,得a3=8.「.a2+a4=20.faiq+faiq+aiq3=2011a3=aiq2=8，Jq=2,解得：o電二2,或11

q=2,1ai=32.又｛an｝為遞增數(shù)列,[q[q=2,la「2.「.a=2%

n(2)vbn=2n-10g22n=-n-2n,/.-Sn=iX2+2X22+3X23+-+nX2n.①「?-2Sn=IX22+2X23+3X24+…+(n-I)X2n+nX2n+i.②①一②得Sn=2+22+23+…+2n—rb2n+i2(i—2n)

i-2-n-2n+i=2n+i-n2+i-2.「S=2n+i-n2+i-2.n4、設(shè)｛an｝是等差數(shù)列，｛乙｝是各項(xiàng)都為正數(shù)的等比數(shù)列，且。1=b1=1,a+b=21,a+b=13

(I)求｛an｝,｛bn｝的通項(xiàng)公式；(II)求數(shù)列1｝＞的前n項(xiàng)和Sn.n解析（［）設(shè)｛,｝的公差為d,｛bj的公比為q解析（［）設(shè)｛,｝的公差為d,｛bj的公比為q，則依題意有q>0且1+4d+q2=13,解得d=2，q=2.所以a=1+(n—1)d=2n—1,b—qnt=2nt.a 2n—1(卬/=7b 2n-1nn21 222n—3 2n—1+ + , 2n-3 2n-2②-②-①得Sn=2+2+-+—+222(1 1=2+2x1+—+—+I2221-工2n-1 2n—1_久 2n+3—2+2x —— =6 TOC\o"1-5"\h\z1 1 2n-1 2n-1 '1——25、已知｛a｝是等差數(shù)列，其前n項(xiàng)和為Sq｛b｝是等比數(shù)列，且a1—b1—2，a4+b4—27,n nn 11 44S4—b4—10.(I)求數(shù)列｛an｝與｛bn｝的通項(xiàng)公式；(II)記T—ab+ab++ab)neN*,證明T+12——2a+10b(neN*).nn1 n—12 1n1 ' n n n解析(1)設(shè)等差數(shù)列｛n｝的公差為d，等比數(shù)列￡n｝的公比為q，由a1—b1—2得

12+3d+2q3=27 Id=34=8+6d，由條件得方程組A+6d—2q3=10,解得|q=2,所以所以an=3n_1,bn(2)證明：由(2)證明：由(1)得Tn=2an+22an—1+…+2na1……①2T=2T=22a+23a+…+2n+1an—1 1②-①得T=—2(3n—1)+3(22+23+…+2n)+2n+2n12(1—2n-1)—( )+2n+2—6n+2=10X2n—6n—101—2而一2a+10b—12=—2(3n—1)+10x2n—12=10x2n—6n—10故T+12=—2a+10b,ngN*.(2012.XX)已知數(shù)列｛an｝的前n項(xiàng)和為Sn,且Sn=2n2+n,nEN*,數(shù)列｛bn｝滿足an=4log2bn+3,nEN*.⑴求an,bn；(2)求數(shù)列｛a；bn｝的前n項(xiàng)和Tn.解析：(1)由Sn=2n2+n,得當(dāng)n=1時(shí)，ai=Si=3；當(dāng)n》2時(shí)，an=Sn-Sn_1=4n-1,所以a=4n-1,nEN*.n由4n-1=an=4log2bn+3,得bn=2n-1,nEN*.(2)由(1)知a；bn=(4n-1)2-1,nEN*,所以Tn=3+7x2+11x22+…+(4n-1)2-1,2Tn=3x2+7x22+…+(4n-5)2-1+(4n-1)2,所以2Tn-Tn=(4n-1)2n-[3+4(2+22+…+2n-1)]

=(4n-5)2n+5.故丁口=(4口-5)2口+5,nEN*.1.(2012?XX)已知數(shù)列色｝的刖n項(xiàng)和Sn=-2n2+kn(其中kEN+),且Sn的最大值為8.⑴確定常數(shù)k,并求an；(2)求數(shù)列9-2an2n［的前n項(xiàng)和匚9=2-n.39=2-n.3.設(shè)｛an｝是等差數(shù)列，｛bn｝是各項(xiàng)都為正數(shù)的等比數(shù)列且a1=b1ia3+b5=21,a5+b3=13.TOC\o"1-5"\h\z一 1解析：(1)當(dāng)n=kEN時(shí)，S=-1n2+kn取最大值，+ n21. . 1.即8=Sk=-2k2+k2=2k2,故如=16，因此壯4,9 7從而a=S-S= n(n》2).又a=S=Z，所以annn-1 2 1 1 2 n9-2an(2)因?yàn)閎=n=9,n2n 2n-123n-1nTn=b1+b2+…+bn=1+2+2；+…+二+2Z，1工。所以T=2T-T=2+1+o+…+o]nnn 2 2n-22n-1(1)求｛an｝,｛bn｝的通項(xiàng)公式；(2)求數(shù)列的前n項(xiàng)和Sn(2)求數(shù)列的前n項(xiàng)和Sn.解析：(1)設(shè)｛an｝的公差為d,｛bn｝的公比為q,則依題意有q>0且，1+2d+q4=21,、1+4d+q2=13,解得彳fd=2,[q=2.所以an=1+(n-l)d=2n-l,bn=qn-i=2n-i.a2n-1(2*==，n3.五+21 222n-32n-22n-12n-12Sn2n-32n-32n-12n-2.②__ 22②-①，得Sn=2+2+2+g+…+2n-12n-22n-1(11 12+2*11+2+五+…+了2n-12n-11-2+2X12n-11-2n-12n-12n+32n-1.4.(2012XX質(zhì)檢)已知數(shù)列｛aj為公差不為零的等差數(shù)列，a「1,各項(xiàng)均為正數(shù)的等比數(shù)列｛bn｝的第1項(xiàng),第3項(xiàng)，第5項(xiàng)分別是a『a3,a21.⑴求數(shù)列｛an｝與｛bj的通項(xiàng)公式；(2)求數(shù)列｛anbn｝的前n項(xiàng)和Sn.解析：⑴設(shè)數(shù)列｛an｝的公差為d(dW0),數(shù)列｛bn｝的公比為q,???由題意得a廠a1a21,.?.(1+2d)2=1X(1+20d),即4d2-16d=0,,.,dW0,，d=4,，an=4n-3./.b1=1,b3=9,b5=81,???｛bn｝的各項(xiàng)均為正數(shù)，，q=3,/.b=3n-1.(2)?.?由(1)可得anbn=(4n-3)3n-1,.?.Sn=30+5X31+9X32+…+(4n-7)X3n-2+(4n-3)X3n-1,3Sn=31+5X32+9X33+…+(4n-7)X3n-1+(4n-3)X3n,兩式相減得：-2Sn=1+4X3+4X32+4X33+…+4X3n-1-(4n-3)X3n

=1+4(3+32+33+…+3”i)—(4n—3)X3n4X3X(1-3n-1)=1+ ' -(4n-3)X3n1-3=(5-4n)X3n-5,.?.S=

n(4n-5)3n+512aM7.（13分）已知數(shù)列｛aj的前n項(xiàng)和Sn與通項(xiàng)an滿足Sn2a