2022屆高考數(shù)學(xué)課標(biāo)版數(shù)學(xué)(文理通用)一輪題型專項(xiàng)練課件-專題4-數(shù)列

專題四數(shù)列專題四數(shù)列-1-4.1數(shù)列小題專項(xiàng)練4.1數(shù)列小題專項(xiàng)練-2--3-1.求數(shù)列通項(xiàng)的常用方法(1)依據(jù)數(shù)列的前幾項(xiàng)求通項(xiàng).(2)由an與Sn的關(guān)系求通項(xiàng).(3)求等差數(shù)列、等比數(shù)列的通項(xiàng),或求可轉(zhuǎn)化為等差數(shù)列、等比數(shù)列的通項(xiàng).2.等差數(shù)列(1)通項(xiàng)公式、等差中項(xiàng)公式、兩種形式的求和公式.(2)常用性質(zhì):①若m+n=p+q(m,n,p,q∈N*),則am+an=ap+aq;②an=am+(n-m)d(m,n∈N*);④已知等差數(shù)列{an},若{an}是遞增數(shù)列,則d>0;若{an}是遞減數(shù)列,則d<0.-3-1.求數(shù)列通項(xiàng)的常用方法④已知等差數(shù)列{an},若{a-4-3.等比數(shù)列(1)通項(xiàng)公式、等比中項(xiàng)公式、公比q=1和q≠1兩種形式的求和公式.(2)常用性質(zhì):①m+n=p+q,則am·an=ap·aq(m,n,p,q∈N*);②an=am·qn-m(m,n∈N*);④已知等比數(shù)列{an},公比q>0,且q≠1.若{an}是遞增數(shù)列,則a1>0,q>1或a1<0,0<q<1;若{an}是遞減數(shù)列,則a1>0,0<q<1或a1<0,q>1.-4-3.等比數(shù)列④已知等比數(shù)列{an},公比q>0,且q≠-5-一二一、選擇題(共12小題,滿分60分)1.記Sn為等差數(shù)列{an}的前n項(xiàng)和,若3S3=S2+S4,a1=2,則a5=(

)A.-12 B.-10 C.10 D.12答案解析解析關(guān)閉因?yàn)?S3=S2+S4,所以3S3=(S3-a3)+(S3+a4),即S3=a4-a3.設(shè)公差為d,則3a1+3d=d,又由a1=2,得d=-3,所以a5=a1+4d=-10.答案解析關(guān)閉B-5-一二一、選擇題(共12小題,滿分60分)答案解析解-6-一二2.等差數(shù)列{an}的首項(xiàng)為1,公差不為0.若a2,a3,a6成等比數(shù)列,則{an}前6項(xiàng)的和為(

)A.-24 B.-3 C.3 D.8答案解析解析關(guān)閉答案解析關(guān)閉-6-一二2.等差數(shù)列{an}的首項(xiàng)為1,公差不為0.若a2-7-一二3.已知等差數(shù)列{an}前9項(xiàng)的和為27,a10=8,則a100=(

)A.100 B.99 C.98 D.97答案解析解析關(guān)閉答案解析關(guān)閉-7-一二3.已知等差數(shù)列{an}前9項(xiàng)的和為27,a10=-8-一二4.已知等差數(shù)列{an}的公差為d,前n項(xiàng)和為Sn,則“d>0”是“S4+S6>2S5”的(

)A.充分不必要條件B.必要不充分條件C.充分必要條件D.既不充分也不必要條件答案解析解析關(guān)閉答案解析關(guān)閉-8-一二4.已知等差數(shù)列{an}的公差為d,前n項(xiàng)和為Sn-9-一二5.記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a4+a5=24,S6=48,則{an}的公差為(

)A.1 B.2 C.4 D.8答案解析解析關(guān)閉答案解析關(guān)閉-9-一二5.記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a4+a-10-一二6.各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和為Sn,若S4=10,S12=130,則S8=(

)A.-30 B.40 C.40或-30 D.40或-50答案解析解析關(guān)閉由等比數(shù)列的性質(zhì),得S4,S8-S4,S12-S8成等比數(shù)列,故(S8-10)2=10×(130-S8),整理可得(S8+30)(S8-40)=0,故S8=40.答案解析關(guān)閉B-10-一二6.各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和為S-11-一二7.已知數(shù)列{an}滿足:=an-1·an+1(n≥2),若a2=3,a2+a4+a6=21,則a4+a6+a8=(

)A.84 B.63 C.42 D.21答案解析解析關(guān)閉答案解析關(guān)閉-11-一二7.已知數(shù)列{an}滿足:=an-1·-12-一二8.已知數(shù)列{an}滿足an+1-an=2,a1=-5,則|a1|+|a2|+…+|a6|=(

)A.9 B.15C.18 D.30答案解析解析關(guān)閉答案解析關(guān)閉-12-一二8.已知數(shù)列{an}滿足an+1-an=2,a1-13-一二9.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an},a5·a6=4,則數(shù)列{log2an}的前10項(xiàng)和為(

)A.5 B.6 C.10 D.12答案解析解析關(guān)閉由等比數(shù)列的性質(zhì)可得a1·a2…·a10=(a1·a10)(a2·a9)…(a5·a6)=(a5·a6)5=45,故log2a1+log2a2+…+log2a10=log2(a1·a2·…·a10)=log245=10,故選C.答案解析關(guān)閉C-13-一二9.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an},a5·a-14-一二10.已知數(shù)列{an}滿足an+1=an-an-1(n≥2),Sn為數(shù)列{an}的前n項(xiàng)和,則S217=(

)A.217a2-a1 B.217a1-a2

C.a1 D.a2答案解析解析關(guān)閉∵an+1=an-an-1(n≥2),∴a3=a2-a1,a4=-a1,a5=-a2,a6=a1-a2,a7=a1,a8=a2,∴數(shù)列{an}的周期為6,S217=S36×6+1=36(a1+a2+a3+a4+a5+a6)+a1=36×0+a1=a1,故選C.答案解析關(guān)閉C-14-一二10.已知數(shù)列{an}滿足an+1=an-an--15-一二答案解析解析關(guān)閉答案解析關(guān)閉-15-一二答案解析解析關(guān)閉答案解析關(guān)閉-16-一二12.(2018浙江,10)已知a1,a2,a3,a4成等比數(shù)列,且a1+a2+a3+a4=ln(a1+a2+a3).若a1>1,則(

)A.a1<a3,a2<a4 B.a1>a3,a2<a4C.a1<a3,a2>a4 D.a1>a3,a2>a4答案解析解析關(guān)閉答案解析關(guān)閉-16-一二12.(2018浙江,10)已知a1,a2,a3-17-一二二、填空題(共4小題,滿分20分)13.記等差數(shù)列{an}的前n項(xiàng)和為Sn,若a3=0,a6+a7=14,則S7=

.

答案解析解析關(guān)閉答案解析關(guān)閉-17-一二二、填空題(共4小題,滿分20分)答案解析解-18-一二14.設(shè)等比數(shù)列{an}滿足a1+a2=-1,a1-a3=-3,則a4=

.

答案解析解析關(guān)閉答案解析關(guān)閉-18-一二14.設(shè)等比數(shù)列{an}滿足a1+a2=-1,a-19-一二15.等差數(shù)列{an}的前n項(xiàng)和為Sn,a3=3,S4=10,則

答案解析解析關(guān)閉答案解析關(guān)閉-19-一二15.等差數(shù)列{an}的前n項(xiàng)和為Sn,a3=3-20-一二16.設(shè)等比數(shù)列{an}滿足a1+a3=10,a2+a4=5,則a1a2…an的最大值為

.

答案解析解析關(guān)閉答案解析關(guān)閉-20-一二16.設(shè)等比數(shù)列{an}滿足a1+a3=10,a4.2數(shù)列大題4.2數(shù)列大題-21--22-1.求通項(xiàng)公式的常見(jiàn)類型(1)已知an與Sn的關(guān)系或Sn與n的關(guān)系,利用公式(2)等差數(shù)列、等比數(shù)列求通項(xiàng)或轉(zhuǎn)化為等差(比)數(shù)列求通項(xiàng).(3)由遞推關(guān)系式求數(shù)列的通項(xiàng)公式.①形如an+1=an+f(n),利用累加法求通項(xiàng).②形如an+1=anf(n),利用累乘法求通項(xiàng).-22-1.求通項(xiàng)公式的常見(jiàn)類型(2)等差數(shù)列、等比數(shù)列求通-23-2.數(shù)列求和的常用方法(1)公式法:利用等差數(shù)列、等比數(shù)列的求和公式.(2)錯(cuò)位相減法:適合求數(shù)列{an·bn}的前n項(xiàng)和Sn,其中{an},{bn}一個(gè)是等差數(shù)列,另一個(gè)是等比數(shù)列.(3)裂項(xiàng)相消法:即將數(shù)列的通項(xiàng)分成兩個(gè)式子的代數(shù)和,通過(guò)累加抵消中間若干項(xiàng)的方法.(4)拆項(xiàng)分組法:先把數(shù)列的每一項(xiàng)拆成兩項(xiàng)(或多項(xiàng)),再重新組合成兩個(gè)(或多個(gè))簡(jiǎn)單的數(shù)列,最后分別求和.(5)并項(xiàng)求和法:把數(shù)列的兩項(xiàng)(或多項(xiàng))組合在一起,重新構(gòu)成一個(gè)數(shù)列再求和,適用于正負(fù)相間排列的數(shù)列求和.3.數(shù)列單調(diào)性的常見(jiàn)題型及方法(1)求最大(小)項(xiàng)時(shí),可利用:①數(shù)列的單調(diào)性;②函數(shù)的單調(diào)性;③導(dǎo)數(shù).(2)求參數(shù)范圍時(shí),可利用:①作差法;②同號(hào)遞推法;③先猜后證法.-23-2.數(shù)列求和的常用方法-24-4.數(shù)列不等式問(wèn)題的解決方法(1)利用數(shù)列(或函數(shù))的單調(diào)性.(2)放縮法:①先求和后放縮;②先放縮后求和,包括放縮后成等差(或等比)數(shù)列再求和,或者放縮后裂項(xiàng)相消再求和.-24-4.數(shù)列不等式問(wèn)題的解決方法4.2.1

等差、等比數(shù)列與數(shù)列

的通項(xiàng)及求和4.2.1等差、等比數(shù)列與數(shù)列

的通項(xiàng)及求和-25--26-考向一考向二考向三考向四考向五等差、等比數(shù)列的通項(xiàng)及求和例1記Sn為等差數(shù)列{an}的前n項(xiàng)和,已知a1=-7,S3=-15.(1)求{an}的通項(xiàng)公式;(2)求Sn,并求Sn的最小值.解:(1)設(shè){an}的公差為d,由題意得3a1+3d=-15.由a1=-7得d=2.所以{an}的通項(xiàng)公式為an=2n-9.(2)由(1)得Sn=n2-8n=(n-4)2-16.所以當(dāng)n=4時(shí),Sn取得最小值,最小值為-16.解題心得對(duì)于等差、等比數(shù)列,求其通項(xiàng)及前n項(xiàng)和時(shí),只需利用等差數(shù)列或等比數(shù)列的通項(xiàng)公式及求和公式求解即可.-26-考向一考向二考向三考向四考向五等差、等比數(shù)列的通項(xiàng)及-27-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練1已知等差數(shù)列{an}的公差不為零,a1=25,且a1,a11,a13成等比數(shù)列.(1)求{an}的通項(xiàng)公式;(2)求a1+a4+a7+…+a3n-2.解:(1)設(shè){an}的公差為d.

即(a1+10d)2=a1(a1+12d).于是d(2a1+25d)=0.又a1=25,所以d=0(舍去)或d=-2.故an=-2n+27.(2)令Sn=a1+a4+a7+…+a3n-2.由(1)知a3n-2=-6n+31,故{a3n-2}是首項(xiàng)為25,公差為-6的等差數(shù)列.-27-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練1已知等差數(shù)列-28-考向一考向二考向三考向四考向五可轉(zhuǎn)化為等差、等比數(shù)列的問(wèn)題例2已知等比數(shù)列{an}的前n項(xiàng)和為Sn,a1=3,且3S1,2S2,S3成等差數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=log3an,求Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1.解:(1)∵3S1,2S2,S3成等差數(shù)列,∴4S2=3S1+S3,∴4(a1+a2)=3a1+(a1+a2+a3),即a3=3a2,∴公比q=3,∴an=a1qn-1=3n.-28-考向一考向二考向三考向四考向五可轉(zhuǎn)化為等差、等比數(shù)列-29-考向一考向二考向三考向四考向五解題心得無(wú)論是求數(shù)列的通項(xiàng)還是求數(shù)列的前n項(xiàng)和,通過(guò)變形、整理后,能夠把數(shù)列轉(zhuǎn)化為等差數(shù)列或等比數(shù)列,進(jìn)而利用等差數(shù)列或等比數(shù)列的通項(xiàng)公式或求和公式解決問(wèn)題.(2)由(1)知,bn=log3an=log33n=n,∵b2n-1b2n-b2nb2n+1=(2n-1)·2n-2n(2n+1)=-4n,∴Tn=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2n-1b2n-b2nb2n+1)-29-考向一考向二考向三考向四考向五解題心得無(wú)論是求數(shù)列的-30-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練2設(shè){an}是公比大于1的等比數(shù)列,Sn為數(shù)列{an}的前n項(xiàng)和,已知S3=7,且a1+3,3a2,a3+4構(gòu)成等差數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;-30-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練2設(shè){an}是-31-考向一考向二考向三考向四考向五解:(1)由已知得

-31-考向一考向二考向三考向四考向五解:(1)由已知得-32-考向一考向二考向三考向四考向五(2)由(1)得a3n+1=23n,∴bn=ln

23n=3nln

2.∵bn+1-bn=3ln

2,∴數(shù)列{bn}為等差數(shù)列.-32-考向一考向二考向三考向四考向五(2)由(1)得a3n-33-考向一考向二考向三考向四考向五求數(shù)列的通項(xiàng)及錯(cuò)位相減求和例3已知{an}為等差數(shù)列,前n項(xiàng)和為Sn(n∈N*),{bn}是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.(1)求{an}和{bn}的通項(xiàng)公式;(2)求數(shù)列{a2nb2n-1}的前n項(xiàng)和(n∈N*).-33-考向一考向二考向三考向四考向五求數(shù)列的通項(xiàng)及錯(cuò)位相減-34-考向一考向二考向三考向四考向五解:(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.由已知b2+b3=12,得b1(q+q2)=12,而b1=2,所以q2+q-6=0.又因?yàn)閝>0,解得q=2.所以,bn=2n.由b3=a4-2a1,可得3d-a1=8.①由S11=11b4,可得a1+5d=16,②聯(lián)立①②,解得a1=1,d=3,由此可得an=3n-2.所以,數(shù)列{an}的通項(xiàng)公式為an=3n-2,數(shù)列{bn}的通項(xiàng)公式為bn=2n.-34-考向一考向二考向三考向四考向五解:(1)設(shè)等差數(shù)列{-35-考向一考向二考向三考向四考向五(2)設(shè)數(shù)列{a2nb2n-1}的前n項(xiàng)和為T(mén)n,由a2n=6n-2,b2n-1=2×4n-1,有a2nb2n-1=(3n-1)×4n,故Tn=2×4+5×42+8×43+…+(3n-1)×4n,4Tn=2×42+5×43+8×44+…+(3n-4)×4n+(3n-1)×4n+1,上述兩式相減,得-3Tn=2×4+3×42+3×43+…+3×4n-(3n-1)×4n+1-35-考向一考向二考向三考向四考向五(2)設(shè)數(shù)列{a2nb-36-考向一考向二考向三考向四考向五解題心得求數(shù)列通項(xiàng)的基本方法是利用等差、等比數(shù)列通項(xiàng)公式,或通過(guò)變形轉(zhuǎn)換成等差、等比數(shù)列求通項(xiàng);如果數(shù)列{an}與數(shù)列{bn}分別是等差數(shù)列和等比數(shù)列,那么數(shù)列{an·bn}的前n項(xiàng)和采用錯(cuò)位相減法來(lái)求.-36-考向一考向二考向三考向四考向五解題心得求數(shù)列通項(xiàng)的基-37-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練3已知等比數(shù)列{an}的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中項(xiàng).數(shù)列{bn}滿足b1=1,數(shù)列{(bn+1-bn)an}的前n項(xiàng)和為2n2+n.(1)求q的值;(2)求數(shù)列{bn}的通項(xiàng)公式.解:(1)由a4+2是a3,a5的等差中項(xiàng),得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.-37-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練3已知等比數(shù)列-38-考向一考向二考向三考向四考向五(2)設(shè)cn=(bn+1-bn)an,數(shù)列{cn}前n項(xiàng)和為Sn,-38-考向一考向二考向三考向四考向五(2)設(shè)cn=(bn+-39-考向一考向二考向三考向四考向五-39-考向一考向二考向三考向四考向五-40-考向一考向二考向三考向四考向五求數(shù)列的通項(xiàng)及裂項(xiàng)求和例4已知數(shù)列{an}的前n項(xiàng)和為Sn,且對(duì)任意正整數(shù)n,都有3an=2Sn+3成立.(1)求數(shù)列{an}的通項(xiàng)公式;解:(1)在3an=2Sn+3中,取n=1,得a1=3,且3an+1=2Sn+1+3,兩式相減,得3an+1-3an=2an+1,∴an+1=3an.∵a1≠0,∴數(shù)列{an}是以3為公比的等比數(shù)列,∴an=3·3n-1=3n.-40-考向一考向二考向三考向四考向五求數(shù)列的通項(xiàng)及裂項(xiàng)求和-41-考向一考向二考向三考向四考向五(2)由(1)得bn=log3an=n,

解題心得對(duì)于已知等式中含有an,Sn的求數(shù)列通項(xiàng)的題目,一般有兩種解題思路,一是消去Sn得到f(an)=0,求出an;二是消去an得到g(Sn)=0,求出Sn,再求an.把數(shù)列的通項(xiàng)拆成兩項(xiàng)之差,求和時(shí)中間的項(xiàng)能夠抵消,從而求得其和.注意抵消后所剩余的項(xiàng)一般前后對(duì)稱.-41-考向一考向二考向三考向四考向五(2)由(1)得bn=-42-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練4Sn為數(shù)列{an}的前n項(xiàng)和.已知an>0,+2an=4Sn+3.(1)求{an}的通項(xiàng)公式;所以{an}是首項(xiàng)為3,公差為2的等差數(shù)列,通項(xiàng)公式為an=2n+1.-42-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練4Sn為數(shù)列{-43-考向一考向二考向三考向四考向五(2)由an=2n+1可知

設(shè)數(shù)列{bn}的前n項(xiàng)和為T(mén)n,則Tn=b1+b2+…+bn-43-考向一考向二考向三考向四考向五(2)由an=2n+1-44-考向一考向二考向三考向四考向五涉及奇偶數(shù)討論的數(shù)列求和例5已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a1=2,S5=30.數(shù)列{bn}的前n項(xiàng)和為T(mén)n,且Tn=2n-1.(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)設(shè)cn=(-1)n(anbn+ln

Sn),求數(shù)列{cn}的前n項(xiàng)和.對(duì)數(shù)列{bn}:當(dāng)n=1時(shí),b1=T1=21-1=1,當(dāng)n≥2時(shí),bn=Tn-Tn-1=2n-2n-1=2n-1,當(dāng)n=1時(shí)也滿足上式.∴bn=2n-1.-44-考向一考向二考向三考向四考向五涉及奇偶數(shù)討論的數(shù)列求-45-考向一考向二考向三考向四考向五(2)cn=(-1)n(anbn+ln

Sn)=(-1)nanbn+(-1)nln

Sn.

∴l(xiāng)n

Sn=lnn(n+1)=lnn+ln(n+1).而(-1)nanbn=(-1)n·2n·2n-1=n·(-2)n,設(shè)數(shù)列{(-1)nanbn}的前n項(xiàng)和為An,數(shù)列{(-1)nln

Sn}的前n項(xiàng)和為Bn,則An=1×(-2)1+2×(-2)2+3×(-2)3+…+n·(-2)n,①則-2An=1×(-2)2+2×(-2)3+3×(-2)4+…+n·(-2)n+1,②①-②得3An=1×(-2)1+(-2)2+(-2)3+…+(-2)n-n·(-2)n+1-45-考向一考向二考向三考向四考向五(2)cn=(-1)n-46-考向一考向二考向三考向四考向五當(dāng)n為偶數(shù)時(shí),Bn=-(ln1+ln2)+(ln2+ln3)-(ln3+ln4)+…+[lnn+ln(n+1)]=ln(n+1)-ln1=ln(n+1);當(dāng)n為奇數(shù)時(shí),Bn=-(ln1+ln2)+(ln2+ln3)-(ln3+ln4)+…-[lnn+ln(n+1)]=-ln(n+1)-ln1=-ln(n+1).由以上可知,Bn=(-1)nln(n+1).-46-考向一考向二考向三考向四考向五當(dāng)n為偶數(shù)時(shí),-47-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練5已知函數(shù)f(x)=4x,4,f(a1),f(a2),…,f(an),2n+3(n∈N*)成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;解:(1)∵4,f(a1),f(a2),…,f(an),2n+3成等比數(shù)列,其公比設(shè)為q,∴2n+3=4×qn+2-1,解得q=2.-47-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練5已知函數(shù)f(-48-考向一考向二考向三考向四考向五-48-考向一考向二考向三考向四考向五4.2.2

數(shù)列中的證明及存在性問(wèn)題4.2.2數(shù)列中的證明及存在性問(wèn)題-49--50-考向一考向二考向三等差(比)數(shù)列的判斷與證明(1)求a1,a2;(2)求數(shù)列{an}的通項(xiàng)公式,并證明數(shù)列{an}是等差數(shù)列;(3)如果數(shù)列{bn}滿足an=log2bn,試證明數(shù)列{bn}是等比數(shù)列,并求其前n項(xiàng)和Tn.又a1=5滿足an=3n+2,所以an=3n+2.因?yàn)閍n+1-an=3(n+1)+2-(3n+2)=3,所以數(shù)列{an}是以5為首項(xiàng),3為公差的等差數(shù)列.-50-考向一考向二考向三等差(比)數(shù)列的判斷與證明(1)求-51-考向一考向二考向三-51-考向一考向二考向三-52-考向一考向二考向三解題心得1.判斷和證明數(shù)列是等差(比)數(shù)列的三種方法.(1)定義法:對(duì)于n≥1的任意自然數(shù),驗(yàn)證an+1-an

為同一常數(shù).(2)通項(xiàng)公式法:若an=kn+b(n∈N*),則{an}為等差數(shù)列;若an=pqkn+b(n∈N*),則{an}為等比數(shù)列.(3)中項(xiàng)公式法:若2an=an-1+an+1(n∈N*,n≥2),則{an}為等差數(shù)列;若

=an-1·an+1(n∈N*,n≥2),則{an}為等比數(shù)列.2.對(duì)已知數(shù)列an與Sn的關(guān)系,證明{an}為等差或等比數(shù)列的問(wèn)題,解題思路是:由an與Sn的關(guān)系遞推出n+1時(shí)的關(guān)系式,兩個(gè)關(guān)系式相減后,進(jìn)行化簡(jiǎn)、整理,最終化歸為用定義法證明.-52-考向一考向二考向三解題心得1.判斷和證明數(shù)列是等差(-53-考向一考向二考向三對(duì)點(diǎn)訓(xùn)練1設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且首項(xiàng)a1≠3,an+1=Sn+3n(n∈N*).(1)求證:{Sn-3n}是等比數(shù)列;(2)若{an}為遞增數(shù)列,求a1的取值范圍.(1)證明:∵an+1=Sn+3n,∴Sn+1=2Sn+3n.∴Sn+1-3n+1=2(Sn-3n).∵a1≠3,∴數(shù)列{Sn-3n}是首項(xiàng)為a1-3,公比為2的等比數(shù)列.-53-考向一考向二考向三對(duì)點(diǎn)訓(xùn)練1設(shè)數(shù)列{an}的前n項(xiàng)和-54-考向一考向二考向三(2)解:由(1)得,Sn-3n=(a1-3)×2n-1,∴Sn=(a1-3)×2n-1+3n.當(dāng)n≥2時(shí),an=Sn-Sn-1=(a1-3)×2n-2+2×3n-1.∵{an}為遞增數(shù)列,∴當(dāng)n≥2時(shí),(a1-3)×2n-1+2×3n>(a1-3)×2n-2+2×3n-1,∵a2=a1+3>a1,∴a1的取值范圍是(-9,+∞).-54-考向一考向二考向三(2)解:由(1)得,Sn-3n=-55-考向一考向二考向三數(shù)列型不等式的證明例2設(shè)Sn是數(shù)列{an}的前n項(xiàng)和,an>0,且4Sn=an(an+2).(1)求數(shù)列{an}的通項(xiàng)公式;(1)解:4Sn=an(an+2),①

即2(an+an-1)=(an+an-1)·(an-an-1).∵an>0,∴an-an-1=2,∴an=2+2(n-1)=2n.-55-考向一考向二考向三數(shù)列型不等式的證明(1)解:4Sn-56-考向一考向二考向三解題心得要證明關(guān)于一個(gè)數(shù)列的前n項(xiàng)和的不等式,一般有兩種思路:一是先求和,再對(duì)和式放縮;二是先對(duì)數(shù)列的通項(xiàng)放縮,再求數(shù)列的和,必要時(shí)對(duì)其和再放縮.-56-考向一考向二考向三解題心得要證明關(guān)于一個(gè)數(shù)列的前n項(xiàng)-57-考向一考向二考向三對(duì)點(diǎn)訓(xùn)練2已知數(shù)列{an}滿足a1=1,an+1=3an+1.-57-考向一考向二考向三對(duì)點(diǎn)訓(xùn)練2已知數(shù)列{an}滿足a1-58-考向一考向二考向三數(shù)列中的存在性問(wèn)題例3已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,an≠0,anan+1=λSn-1,其中λ為常數(shù).(1)證明:an+2-an=λ;(2)是否存在λ,使得{an}為等差數(shù)列?并說(shuō)明理由.(1)證明:由題設(shè),anan+1=λSn-1,an+1an+2=λSn+1-1,兩式相減,得an+1(an+2-an)=λan+1.因?yàn)閍n+1≠0,所以an+2-an=λ.-58-考向一考向二考向三數(shù)列中的存在性問(wèn)題(1)證明:由題-59-考向一考向二考向三解題心得假設(shè)推理法:先假設(shè)所探求對(duì)象存在或結(jié)論成立,以此假設(shè)為前提條件進(jìn)行運(yùn)算或邏輯推理,若由此推出矛盾,則假設(shè)不成立,即不存在.若推不出矛盾,即得到存在的結(jié)果.(2)解:由題設(shè),a1=1,a1a2=λS1-1,可得a2=λ-1.由(1)知,a3=λ+1.令2a2=a1+a3,解得λ=4.故an+2-an=4.由此可得{a2n-1}是首項(xiàng)為1,公差為4的等差數(shù)列,a2n-1=4n-3;{a2n}是首項(xiàng)為3,公差為4的等差數(shù)列,a2n=4n-1.所以an=2n-1,an+1-an=2.因此存在λ=4,使得數(shù)列{an}為等差數(shù)列.-59-考向一考向二考向三解題心得假設(shè)推理法:先假設(shè)所探求對(duì)-60-考向一考向二考向三對(duì)點(diǎn)訓(xùn)練3已知數(shù)列{an}和{bn},a1a2a3…an=(n∈N*),且a1=2,b3-b2=3,數(shù)列{an}為等比數(shù)列,公比為q.(1)求a3及數(shù)列{bn}的通項(xiàng)公式;(2)令cn=,是否存在正整數(shù)m,n(m≠n),使c2,cm,cn成等差數(shù)列?若存在,求出m,n的值;若不存在,請(qǐng)說(shuō)明理由.-60-考向一考向二考向三對(duì)點(diǎn)訓(xùn)練3已知數(shù)列{an}和{bn-61-考向一考向二考向三-61-考向一考向二考向三專題四數(shù)列專題四數(shù)列-62-4.1數(shù)列小題專項(xiàng)練4.1數(shù)列小題專項(xiàng)練-63--64-1.求數(shù)列通項(xiàng)的常用方法(1)依據(jù)數(shù)列的前幾項(xiàng)求通項(xiàng).(2)由an與Sn的關(guān)系求通項(xiàng).(3)求等差數(shù)列、等比數(shù)列的通項(xiàng),或求可轉(zhuǎn)化為等差數(shù)列、等比數(shù)列的通項(xiàng).2.等差數(shù)列(1)通項(xiàng)公式、等差中項(xiàng)公式、兩種形式的求和公式.(2)常用性質(zhì):①若m+n=p+q(m,n,p,q∈N*),則am+an=ap+aq;②an=am+(n-m)d(m,n∈N*);④已知等差數(shù)列{an},若{an}是遞增數(shù)列,則d>0;若{an}是遞減數(shù)列,則d<0.-3-1.求數(shù)列通項(xiàng)的常用方法④已知等差數(shù)列{an},若{a-65-3.等比數(shù)列(1)通項(xiàng)公式、等比中項(xiàng)公式、公比q=1和q≠1兩種形式的求和公式.(2)常用性質(zhì):①m+n=p+q,則am·an=ap·aq(m,n,p,q∈N*);②an=am·qn-m(m,n∈N*);④已知等比數(shù)列{an},公比q>0,且q≠1.若{an}是遞增數(shù)列,則a1>0,q>1或a1<0,0<q<1;若{an}是遞減數(shù)列,則a1>0,0<q<1或a1<0,q>1.-4-3.等比數(shù)列④已知等比數(shù)列{an},公比q>0,且q≠-66-一二一、選擇題(共12小題,滿分60分)1.記Sn為等差數(shù)列{an}的前n項(xiàng)和,若3S3=S2+S4,a1=2,則a5=(

)A.-12 B.-10 C.10 D.12答案解析解析關(guān)閉因?yàn)?S3=S2+S4,所以3S3=(S3-a3)+(S3+a4),即S3=a4-a3.設(shè)公差為d,則3a1+3d=d,又由a1=2,得d=-3,所以a5=a1+4d=-10.答案解析關(guān)閉B-5-一二一、選擇題(共12小題,滿分60分)答案解析解-67-一二2.等差數(shù)列{an}的首項(xiàng)為1,公差不為0.若a2,a3,a6成等比數(shù)列,則{an}前6項(xiàng)的和為(

)A.-24 B.-3 C.3 D.8答案解析解析關(guān)閉答案解析關(guān)閉-6-一二2.等差數(shù)列{an}的首項(xiàng)為1,公差不為0.若a2-68-一二3.已知等差數(shù)列{an}前9項(xiàng)的和為27,a10=8,則a100=(

)A.100 B.99 C.98 D.97答案解析解析關(guān)閉答案解析關(guān)閉-7-一二3.已知等差數(shù)列{an}前9項(xiàng)的和為27,a10=-69-一二4.已知等差數(shù)列{an}的公差為d,前n項(xiàng)和為Sn,則“d>0”是“S4+S6>2S5”的(

)A.充分不必要條件B.必要不充分條件C.充分必要條件D.既不充分也不必要條件答案解析解析關(guān)閉答案解析關(guān)閉-8-一二4.已知等差數(shù)列{an}的公差為d,前n項(xiàng)和為Sn-70-一二5.記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a4+a5=24,S6=48,則{an}的公差為(

)A.1 B.2 C.4 D.8答案解析解析關(guān)閉答案解析關(guān)閉-9-一二5.記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a4+a-71-一二6.各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和為Sn,若S4=10,S12=130,則S8=(

)A.-30 B.40 C.40或-30 D.40或-50答案解析解析關(guān)閉由等比數(shù)列的性質(zhì),得S4,S8-S4,S12-S8成等比數(shù)列,故(S8-10)2=10×(130-S8),整理可得(S8+30)(S8-40)=0,故S8=40.答案解析關(guān)閉B-10-一二6.各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和為S-72-一二7.已知數(shù)列{an}滿足:=an-1·an+1(n≥2),若a2=3,a2+a4+a6=21,則a4+a6+a8=(

)A.84 B.63 C.42 D.21答案解析解析關(guān)閉答案解析關(guān)閉-11-一二7.已知數(shù)列{an}滿足:=an-1·-73-一二8.已知數(shù)列{an}滿足an+1-an=2,a1=-5,則|a1|+|a2|+…+|a6|=(

)A.9 B.15C.18 D.30答案解析解析關(guān)閉答案解析關(guān)閉-12-一二8.已知數(shù)列{an}滿足an+1-an=2,a1-74-一二9.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an},a5·a6=4,則數(shù)列{log2an}的前10項(xiàng)和為(

)A.5 B.6 C.10 D.12答案解析解析關(guān)閉由等比數(shù)列的性質(zhì)可得a1·a2…·a10=(a1·a10)(a2·a9)…(a5·a6)=(a5·a6)5=45,故log2a1+log2a2+…+log2a10=log2(a1·a2·…·a10)=log245=10,故選C.答案解析關(guān)閉C-13-一二9.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an},a5·a-75-一二10.已知數(shù)列{an}滿足an+1=an-an-1(n≥2),Sn為數(shù)列{an}的前n項(xiàng)和,則S217=(

)A.217a2-a1 B.217a1-a2

C.a1 D.a2答案解析解析關(guān)閉∵an+1=an-an-1(n≥2),∴a3=a2-a1,a4=-a1,a5=-a2,a6=a1-a2,a7=a1,a8=a2,∴數(shù)列{an}的周期為6,S217=S36×6+1=36(a1+a2+a3+a4+a5+a6)+a1=36×0+a1=a1,故選C.答案解析關(guān)閉C-14-一二10.已知數(shù)列{an}滿足an+1=an-an--76-一二答案解析解析關(guān)閉答案解析關(guān)閉-15-一二答案解析解析關(guān)閉答案解析關(guān)閉-77-一二12.(2018浙江,10)已知a1,a2,a3,a4成等比數(shù)列,且a1+a2+a3+a4=ln(a1+a2+a3).若a1>1,則(

)A.a1<a3,a2<a4 B.a1>a3,a2<a4C.a1<a3,a2>a4 D.a1>a3,a2>a4答案解析解析關(guān)閉答案解析關(guān)閉-16-一二12.(2018浙江,10)已知a1,a2,a3-78-一二二、填空題(共4小題,滿分20分)13.記等差數(shù)列{an}的前n項(xiàng)和為Sn,若a3=0,a6+a7=14,則S7=

.

答案解析解析關(guān)閉答案解析關(guān)閉-17-一二二、填空題(共4小題,滿分20分)答案解析解-79-一二14.設(shè)等比數(shù)列{an}滿足a1+a2=-1,a1-a3=-3,則a4=

.

答案解析解析關(guān)閉答案解析關(guān)閉-18-一二14.設(shè)等比數(shù)列{an}滿足a1+a2=-1,a-80-一二15.等差數(shù)列{an}的前n項(xiàng)和為Sn,a3=3,S4=10,則

答案解析解析關(guān)閉答案解析關(guān)閉-19-一二15.等差數(shù)列{an}的前n項(xiàng)和為Sn,a3=3-81-一二16.設(shè)等比數(shù)列{an}滿足a1+a3=10,a2+a4=5,則a1a2…an的最大值為

.

答案解析解析關(guān)閉答案解析關(guān)閉-20-一二16.設(shè)等比數(shù)列{an}滿足a1+a3=10,a4.2數(shù)列大題4.2數(shù)列大題-82--83-1.求通項(xiàng)公式的常見(jiàn)類型(1)已知an與Sn的關(guān)系或Sn與n的關(guān)系,利用公式(2)等差數(shù)列、等比數(shù)列求通項(xiàng)或轉(zhuǎn)化為等差(比)數(shù)列求通項(xiàng).(3)由遞推關(guān)系式求數(shù)列的通項(xiàng)公式.①形如an+1=an+f(n),利用累加法求通項(xiàng).②形如an+1=anf(n),利用累乘法求通項(xiàng).-22-1.求通項(xiàng)公式的常見(jiàn)類型(2)等差數(shù)列、等比數(shù)列求通-84-2.數(shù)列求和的常用方法(1)公式法:利用等差數(shù)列、等比數(shù)列的求和公式.(2)錯(cuò)位相減法:適合求數(shù)列{an·bn}的前n項(xiàng)和Sn,其中{an},{bn}一個(gè)是等差數(shù)列,另一個(gè)是等比數(shù)列.(3)裂項(xiàng)相消法:即將數(shù)列的通項(xiàng)分成兩個(gè)式子的代數(shù)和,通過(guò)累加抵消中間若干項(xiàng)的方法.(4)拆項(xiàng)分組法:先把數(shù)列的每一項(xiàng)拆成兩項(xiàng)(或多項(xiàng)),再重新組合成兩個(gè)(或多個(gè))簡(jiǎn)單的數(shù)列,最后分別求和.(5)并項(xiàng)求和法:把數(shù)列的兩項(xiàng)(或多項(xiàng))組合在一起,重新構(gòu)成一個(gè)數(shù)列再求和,適用于正負(fù)相間排列的數(shù)列求和.3.數(shù)列單調(diào)性的常見(jiàn)題型及方法(1)求最大(小)項(xiàng)時(shí),可利用:①數(shù)列的單調(diào)性;②函數(shù)的單調(diào)性;③導(dǎo)數(shù).(2)求參數(shù)范圍時(shí),可利用:①作差法;②同號(hào)遞推法;③先猜后證法.-23-2.數(shù)列求和的常用方法-85-4.數(shù)列不等式問(wèn)題的解決方法(1)利用數(shù)列(或函數(shù))的單調(diào)性.(2)放縮法:①先求和后放縮;②先放縮后求和,包括放縮后成等差(或等比)數(shù)列再求和,或者放縮后裂項(xiàng)相消再求和.-24-4.數(shù)列不等式問(wèn)題的解決方法4.2.1

等差、等比數(shù)列與數(shù)列

的通項(xiàng)及求和4.2.1等差、等比數(shù)列與數(shù)列

的通項(xiàng)及求和-86--87-考向一考向二考向三考向四考向五等差、等比數(shù)列的通項(xiàng)及求和例1記Sn為等差數(shù)列{an}的前n項(xiàng)和,已知a1=-7,S3=-15.(1)求{an}的通項(xiàng)公式;(2)求Sn,并求Sn的最小值.解:(1)設(shè){an}的公差為d,由題意得3a1+3d=-15.由a1=-7得d=2.所以{an}的通項(xiàng)公式為an=2n-9.(2)由(1)得Sn=n2-8n=(n-4)2-16.所以當(dāng)n=4時(shí),Sn取得最小值,最小值為-16.解題心得對(duì)于等差、等比數(shù)列,求其通項(xiàng)及前n項(xiàng)和時(shí),只需利用等差數(shù)列或等比數(shù)列的通項(xiàng)公式及求和公式求解即可.-26-考向一考向二考向三考向四考向五等差、等比數(shù)列的通項(xiàng)及-88-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練1已知等差數(shù)列{an}的公差不為零,a1=25,且a1,a11,a13成等比數(shù)列.(1)求{an}的通項(xiàng)公式;(2)求a1+a4+a7+…+a3n-2.解:(1)設(shè){an}的公差為d.

即(a1+10d)2=a1(a1+12d).于是d(2a1+25d)=0.又a1=25,所以d=0(舍去)或d=-2.故an=-2n+27.(2)令Sn=a1+a4+a7+…+a3n-2.由(1)知a3n-2=-6n+31,故{a3n-2}是首項(xiàng)為25,公差為-6的等差數(shù)列.-27-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練1已知等差數(shù)列-89-考向一考向二考向三考向四考向五可轉(zhuǎn)化為等差、等比數(shù)列的問(wèn)題例2已知等比數(shù)列{an}的前n項(xiàng)和為Sn,a1=3,且3S1,2S2,S3成等差數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=log3an,求Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1.解:(1)∵3S1,2S2,S3成等差數(shù)列,∴4S2=3S1+S3,∴4(a1+a2)=3a1+(a1+a2+a3),即a3=3a2,∴公比q=3,∴an=a1qn-1=3n.-28-考向一考向二考向三考向四考向五可轉(zhuǎn)化為等差、等比數(shù)列-90-考向一考向二考向三考向四考向五解題心得無(wú)論是求數(shù)列的通項(xiàng)還是求數(shù)列的前n項(xiàng)和,通過(guò)變形、整理后,能夠把數(shù)列轉(zhuǎn)化為等差數(shù)列或等比數(shù)列,進(jìn)而利用等差數(shù)列或等比數(shù)列的通項(xiàng)公式或求和公式解決問(wèn)題.(2)由(1)知,bn=log3an=log33n=n,∵b2n-1b2n-b2nb2n+1=(2n-1)·2n-2n(2n+1)=-4n,∴Tn=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2n-1b2n-b2nb2n+1)-29-考向一考向二考向三考向四考向五解題心得無(wú)論是求數(shù)列的-91-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練2設(shè){an}是公比大于1的等比數(shù)列,Sn為數(shù)列{an}的前n項(xiàng)和,已知S3=7,且a1+3,3a2,a3+4構(gòu)成等差數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;-30-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練2設(shè){an}是-92-考向一考向二考向三考向四考向五解:(1)由已知得

-31-考向一考向二考向三考向四考向五解:(1)由已知得-93-考向一考向二考向三考向四考向五(2)由(1)得a3n+1=23n,∴bn=ln

23n=3nln

2.∵bn+1-bn=3ln

2,∴數(shù)列{bn}為等差數(shù)列.-32-考向一考向二考向三考向四考向五(2)由(1)得a3n-94-考向一考向二考向三考向四考向五求數(shù)列的通項(xiàng)及錯(cuò)位相減求和例3已知{an}為等差數(shù)列,前n項(xiàng)和為Sn(n∈N*),{bn}是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.(1)求{an}和{bn}的通項(xiàng)公式;(2)求數(shù)列{a2nb2n-1}的前n項(xiàng)和(n∈N*).-33-考向一考向二考向三考向四考向五求數(shù)列的通項(xiàng)及錯(cuò)位相減-95-考向一考向二考向三考向四考向五解:(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.由已知b2+b3=12,得b1(q+q2)=12,而b1=2,所以q2+q-6=0.又因?yàn)閝>0,解得q=2.所以,bn=2n.由b3=a4-2a1,可得3d-a1=8.①由S11=11b4,可得a1+5d=16,②聯(lián)立①②,解得a1=1,d=3,由此可得an=3n-2.所以,數(shù)列{an}的通項(xiàng)公式為an=3n-2,數(shù)列{bn}的通項(xiàng)公式為bn=2n.-34-考向一考向二考向三考向四考向五解:(1)設(shè)等差數(shù)列{-96-考向一考向二考向三考向四考向五(2)設(shè)數(shù)列{a2nb2n-1}的前n項(xiàng)和為T(mén)n,由a2n=6n-2,b2n-1=2×4n-1,有a2nb2n-1=(3n-1)×4n,故Tn=2×4+5×42+8×43+…+(3n-1)×4n,4Tn=2×42+5×43+8×44+…+(3n-4)×4n+(3n-1)×4n+1,上述兩式相減,得-3Tn=2×4+3×42+3×43+…+3×4n-(3n-1)×4n+1-35-考向一考向二考向三考向四考向五(2)設(shè)數(shù)列{a2nb-97-考向一考向二考向三考向四考向五解題心得求數(shù)列通項(xiàng)的基本方法是利用等差、等比數(shù)列通項(xiàng)公式,或通過(guò)變形轉(zhuǎn)換成等差、等比數(shù)列求通項(xiàng);如果數(shù)列{an}與數(shù)列{bn}分別是等差數(shù)列和等比數(shù)列,那么數(shù)列{an·bn}的前n項(xiàng)和采用錯(cuò)位相減法來(lái)求.-36-考向一考向二考向三考向四考向五解題心得求數(shù)列通項(xiàng)的基-98-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練3已知等比數(shù)列{an}的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中項(xiàng).數(shù)列{bn}滿足b1=1,數(shù)列{(bn+1-bn)an}的前n項(xiàng)和為2n2+n.(1)求q的值;(2)求數(shù)列{bn}的通項(xiàng)公式.解:(1)由a4+2是a3,a5的等差中項(xiàng),得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.-37-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練3已知等比數(shù)列-99-考向一考向二考向三考向四考向五(2)設(shè)cn=(bn+1-bn)an,數(shù)列{cn}前n項(xiàng)和為Sn,-38-考向一考向二考向三考向四考向五(2)設(shè)cn=(bn+-100-考向一考向二考向三考向四考向五-39-考向一考向二考向三考向四考向五-101-考向一考向二考向三考向四考向五求數(shù)列的通項(xiàng)及裂項(xiàng)求和例4已知數(shù)列{an}的前n項(xiàng)和為Sn,且對(duì)任意正整數(shù)n,都有3an=2Sn+3成立.(1)求數(shù)列{an}的通項(xiàng)公式;解:(1)在3an=2Sn+3中,取n=1,得a1=3,且3an+1=2Sn+1+3,兩式相減,得3an+1-3an=2an+1,∴an+1=3an.∵a1≠0,∴數(shù)列{an}是以3為公比的等比數(shù)列,∴an=3·3n-1=3n.-40-考向一考向二考向三考向四考向五求數(shù)列的通項(xiàng)及裂項(xiàng)求和-102-考向一考向二考向三考向四考向五(2)由(1)得bn=log3an=n,

解題心得對(duì)于已知等式中含有an,Sn的求數(shù)列通項(xiàng)的題目,一般有兩種解題思路,一是消去Sn得到f(an)=0,求出an;二是消去an得到g(Sn)=0,求出Sn,再求an.把數(shù)列的通項(xiàng)拆成兩項(xiàng)之差,求和時(shí)中間的項(xiàng)能夠抵消,從而求得其和.注意抵消后所剩余的項(xiàng)一般前后對(duì)稱.-41-考向一考向二考向三考向四考向五(2)由(1)得bn=-103-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練4Sn為數(shù)列{an}的前n項(xiàng)和.已知an>0,+2an=4Sn+3.(1)求{an}的通項(xiàng)公式;所以{an}是首項(xiàng)為3,公差為2的等差數(shù)列,通項(xiàng)公式為an=2n+1.-42-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練4Sn為數(shù)列{-104-考向一考向二考向三考向四考向五(2)由an=2n+1可知

設(shè)數(shù)列{bn}的前n項(xiàng)和為T(mén)n,則Tn=b1+b2+…+bn-43-考向一考向二考向三考向四考向五(2)由an=2n+1-105-考向一考向二考向三考向四考向五涉及奇偶數(shù)討論的數(shù)列求和例5已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a1=2,S5=30.數(shù)列{bn}的前n項(xiàng)和為T(mén)n,且Tn=2n-1.(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)設(shè)cn=(-1)n(anbn+ln

Sn),求數(shù)列{cn}的前n項(xiàng)和.對(duì)數(shù)列{bn}:當(dāng)n=1時(shí),b1=T1=21-1=1,當(dāng)n≥2時(shí),bn=Tn-Tn-1=2n-2n-1=2n-1,當(dāng)n=1時(shí)也滿足上式.∴bn=2n-1.-44-考向一考向二考向三考向四考向五涉及奇偶數(shù)討論的數(shù)列求-106-考向一考向二考向三考向四考向五(2)cn=(-1)n(anbn+ln

Sn)=(-1)nanbn+(-1)nln

Sn.

∴l(xiāng)n

Sn=lnn(n+1)=lnn+ln(n+1).而(-1)nanbn=(-1)n·2n·2n-1=n·(-2)n,設(shè)數(shù)列{(-1)nanbn}的前n項(xiàng)和為An,數(shù)列{(-1)nln

Sn}的前n項(xiàng)和為Bn,則An=1×(-2)1+2×(-2)2+3×(-2)3+…+n·(-2)n,①則-2An=1×(-2)2+2×(-2)3+3×(-2)4+…+n·(-2)n+1,②①-②得3An=1×(-2)1+(-2)2+(-2)3+…+(-2)n-n·(-2)n+1-45-考向一考向二考向三考向四考向五(2)cn=(-1)n-107-考向一考向二考向三考向四考向五當(dāng)n為偶數(shù)時(shí),Bn=-(ln1+ln2)+(ln2+ln3)-(ln3+ln4)+…+[lnn+ln(n+1)]=ln(n+1)-ln1=ln(n+1);當(dāng)n為奇數(shù)時(shí),Bn=-(ln1+ln2)+(ln2+ln3)-(ln3+ln4)+…-[lnn+ln(n+1)]=-ln(n+1)-ln1=-ln(n+1).由以上可知,Bn=(-1)nln(n+1).-46-考向一考向二考向三考向四考向五當(dāng)n為偶數(shù)時(shí),-108-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練5已知函數(shù)f(x)=4x,4,f(a1),f(a2),…,f(an),2n+3(n∈N*)成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;解:(1)∵4,f(a1),f(a2),…,f(an),2n+3成等比數(shù)列,其公比設(shè)為q,∴2n+3=4×qn+2-1,解得q=2.-47-考向一考向二考向三考向四考向五對(duì)點(diǎn)訓(xùn)練5已知函數(shù)f(-109-考向一考向二考向三考向四考向五-48-考向一考向二考向三考向四考向五4.2.2

數(shù)列中的證明及存在性問(wèn)題4.2.2數(shù)列中的證明及存在性問(wèn)題-110--111-考向一考向二考向三等差(比)數(shù)列的判斷與證明(1)求a1,a2;(2)求數(shù)列{an}的通項(xiàng)公式,并證明數(shù)列{an}是等差數(shù)列;(3)如果數(shù)列{bn}滿足an=log2bn,試證明數(shù)列{bn}是等比數(shù)列,并求其前n項(xiàng)和Tn.又a1=5滿足an=3n+2,所以an=3n+2.因?yàn)閍n+1-an=3(n+1)+2-(3n+2)=3,所以數(shù)列{an}是以5為首項(xiàng),3