202X屆高考數(shù)學(xué)一輪復(fù)習(xí)第六章數(shù)列6.4數(shù)列求和、數(shù)列的綜合應(yīng)用課件文

1、6.4數(shù)列求和、數(shù)列的綜合應(yīng)用高考文數(shù)高考文數(shù) (課標(biāo)專用)1.(2017課標(biāo)全國(guó),17,12分)設(shè)數(shù)列an滿足a1+3a2+(2n-1)an=2n.(1)求an的通項(xiàng)公式;(2)求數(shù)列的前n項(xiàng)和.21nan五年高考A A組組 統(tǒng)一命題統(tǒng)一命題課標(biāo)卷題組課標(biāo)卷題組解析解析(1)因?yàn)閍1+3a2+(2n-1)an=2n,故當(dāng)n2時(shí),a1+3a2+(2n-3)an-1=2(n-1).兩式相減得(2n-1)an=2.所以an=(n2).又由題設(shè)可得a1=2,適合上式,從而an的通項(xiàng)公式為an=(nN*).(2)記的前n項(xiàng)和為Sn.由(1)知=-.則Sn=-+-+-=.221n221n21nan21n

2、an2(21)(21)nn121n121n11131315121n121n221nn思路分析思路分析(1)條件a1+3a2+(2n-1)an=2n的實(shí)質(zhì)就是數(shù)列(2n-1)an的前n項(xiàng)和,故可利用an與Sn的關(guān)系求解.(2)利用(1)的結(jié)果求得的通項(xiàng)公式,然后用裂項(xiàng)相消法求和.21nan易錯(cuò)警示易錯(cuò)警示(1)要注意n=1時(shí),是否符合所求得的通項(xiàng)公式;(2)裂項(xiàng)相消后,注意留下了哪些項(xiàng),避免遺漏.2.(2016課標(biāo)全國(guó),17,12分)等差數(shù)列an中,a3+a4=4,a5+a7=6.(1)求an的通項(xiàng)公式;(2)設(shè)bn=an,求數(shù)列bn的前10項(xiàng)和,其中x表示不超過(guò)x的最大整數(shù),如0.9=0,2.

3、6=2.解析解析(1)設(shè)數(shù)列an的公差為d,由題意有解得(3分)所以an的通項(xiàng)公式為an=.(5分)(2)由(1)知,bn=.(6分)當(dāng)n=1,2,3時(shí),12,bn=1;當(dāng)n=4,5時(shí),23,bn=2;當(dāng)n=6,7,8時(shí),34,bn=3;當(dāng)n=9,10時(shí),40,可得q=2,故bn=2n-1.所以Tn=2n-1.設(shè)等差數(shù)列an的公差為d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,從而a1=1,d=1,故an=n,所以Sn=.(2)由(1),有T1+T2+Tn=(21+22+2n)-n=-n=2n+1-n-2.由Sn+(T1+T2+Tn)=an+4bn

4、可得+2n+1-n-2=n+2n+1,整理得n2-3n-4=0,解得n=-1(舍)或n=4.所以n的值為4.1212n(1)2n n2 (12 )12n(1)2n n2.(2017山東,19,12分)已知an是各項(xiàng)均為正數(shù)的等比數(shù)列,且a1+a2=6,a1a2=a3.(1)求數(shù)列an的通項(xiàng)公式;(2)bn為各項(xiàng)非零的等差數(shù)列,其前n項(xiàng)和為Sn.已知S2n+1=bnbn+1,求數(shù)列的前n項(xiàng)和Tn.nnba解析解析(1)設(shè)an的公比為q,由題意知:a1(1+q)=6,q=a1q2,又an0,解得a1=2,q=2,所以an=2n.(2)由題意知:S2n+1=(2n+1)bn+1,又S2n+1=bnb

5、n+1,bn+10,所以bn=2n+1.令cn=,則cn=.因此Tn=c1+c2+cn=+,又Tn=+,兩式相減得Tn=+-,所以Tn=5-.21a121(21)()2nnbbnnba212nn322523721212nn212nn12232352472212nn1212nn123221111222n1212nn252nn3.(2015安徽,18,12分)已知數(shù)列an是遞增的等比數(shù)列,且a1+a4=9,a2a3=8.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)Sn為數(shù)列an的前n項(xiàng)和,bn=,求數(shù)列bn的前n項(xiàng)和Tn.11nnnaS S解析解析(1)由題設(shè)知a1a4=a2a3=8,又a1+a4=9,可

6、解得或(舍去).由a4=a1q3得公比為q=2,故an=a1qn-1=2n-1.(2)Sn=2n-1,又bn=-,所以Tn=b1+b2+bn=+=-=1-.141,8aa148,1aa1(1)1naqq11nnnaS S11nnnnSSS S1nS11nS1211SS2311SS111nnSS11S11nS1121n評(píng)析評(píng)析本題考查等比數(shù)列通項(xiàng)公式及等比數(shù)列性質(zhì),等比數(shù)列求和.考點(diǎn)二數(shù)列的綜合應(yīng)用考點(diǎn)二數(shù)列的綜合應(yīng)用1.(2015福建,16,4分)若a,b是函數(shù)f(x)=x2-px+q(p0,q0)的兩個(gè)不同的零點(diǎn),且a,b,-2這三個(gè)數(shù)可適當(dāng)排序后成等差數(shù)列,也可適當(dāng)排序后成等比數(shù)列,則p+

7、q的值等于.答案答案9解析解析依題意有a,b是方程x2-px+q=0的兩根,則a+b=p,ab=q,由p0,q0可知a0,b0.由題意可知ab=(-2)2=4=q,a-2=2b或b-2=2a,將a-2=2b代入ab=4可解得a=4,b=1,此時(shí)a+b=5,將b-2=2a代入ab=4可解得a=1,b=4,此時(shí)a+b=5,則p=5,q=4,故p+q=9.2.(2019天津,18,13分)設(shè)an是等差數(shù)列,bn是等比數(shù)列,公比大于0.已知a1=b1=3,b2=a3,b3=4a2+3.(1)求an和bn的通項(xiàng)公式;(2)設(shè)數(shù)列cn滿足cn=求a1c1+a2c2+a2nc2n(nN*).21,.nnbn

8、為奇數(shù)為偶數(shù)解析解析本小題主要考查等差數(shù)列、等比數(shù)列的通項(xiàng)公式及其前n項(xiàng)和公式等基礎(chǔ)知識(shí).考查數(shù)列求和的基本方法和運(yùn)算求解能力,體現(xiàn)了數(shù)學(xué)運(yùn)算素養(yǎng).(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.依題意,得解得故an=3+3(n-1)=3n,bn=33n-1=3n.所以,an的通項(xiàng)公式為an=3n,bn的通項(xiàng)公式為bn=3n.(2)a1c1+a2c2+a2nc2n=(a1+a3+a5+a2n-1)+(a2b1+a4b2+a6b3+a2nbn)=+(631+1232+1833+6n3n)=3n2+6(131+232+n3n).2332 ,3154 ,qdqd3,3,dq(1)362n n

9、n 記Tn=131+232+n3n,則3Tn=132+233+n3n+1,-得,2Tn=-3-32-33-3n+n3n+1=-+n3n+1=.3(1 3 )1 3n1(21)332nn所以,a1c1+a2c2+a2nc2n=3n2+6Tn=3n2+3=(nN*).1(21)332nn22(21)3692nnn思路分析思路分析(1)利用等差、等比數(shù)列的通項(xiàng)公式求出公差d,公比q即可.(2)利用cn的通項(xiàng)公式,進(jìn)行分組求和,在計(jì)算差比數(shù)列時(shí)采用錯(cuò)位相減法求和.解題關(guān)鍵解題關(guān)鍵根據(jù)n的奇偶性得數(shù)列cn的通項(xiàng)公式,從而選擇合適的求和方法是求解的關(guān)鍵.3.(2018浙江,20,15分)已知等比數(shù)列an的

10、公比q1,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.解析解析本題主要考查等差數(shù)列、等比數(shù)列、數(shù)列求和等基礎(chǔ)知識(shí),同時(shí)考查運(yùn)算求解能力和綜合應(yīng)用能力.(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.由a3+a5=20得8=20,解得q=2或q=,因?yàn)閝1,所以q=2.(2)設(shè)cn=(bn+1-bn)an,數(shù)列cn的前n項(xiàng)和為Sn.由cn=解得cn=4n-1.由(1)可知an=2n-1,所以bn+1

11、-bn=(4n-1),故bn-bn-1=(4n-5),n2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+(b3-b2)+(b2-b1)1qq1211,1,2,nnSnSSn112n212n=(4n-5)+(4n-9)+7+3.設(shè)Tn=3+7+11+(4n-5),n2,Tn=3+7+(4n-9)+(4n-5),所以Tn=3+4+4+4-(4n-5),因此Tn=14-(4n+3),n2,又b1=1,所以bn=15-(4n+3).212n312n1212212212n1212212212n112n1212212212n112n212n212n易錯(cuò)警示易錯(cuò)警示利用錯(cuò)位相減法求和時(shí),要注意以

12、下幾點(diǎn):(1)錯(cuò)位相減法求和,只適合于數(shù)列anbn,其中an為等差數(shù)列,bn為等比數(shù)列.(2)在等式兩邊所乘的數(shù)是等比數(shù)列bn的公比.(3)兩式相減時(shí),一定要錯(cuò)開(kāi)一位.4.(2017天津,18,13分)已知an為等差數(shù)列,前n項(xiàng)和為Sn(nN*),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a2nbn的前n項(xiàng)和(nN*).解析解析(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

13、=2n.由b3=a4-2a1,可得3d-a1=8.由S11=11b4,可得a1+5d=16,聯(lián)立,解得a1=1,d=3,由此可得an=3n-2.所以,an的通項(xiàng)公式為an=3n-2,bn的通項(xiàng)公式為bn=2n.(2)設(shè)數(shù)列a2nbn的前n項(xiàng)和為Tn,由a2n=6n-2,有Tn=42+1022+1623+(6n-2)2n,2Tn=422+1023+1624+(6n-8)2n+(6n-2)2n+1,上述兩式相減,得-Tn=42+622+623+62n-(6n-2)2n+1=-4-(6n-2)2n+1=-(3n-4)2n+2-16.12 (12 )12n得Tn=(3n-4)2n+2+16.所以,數(shù)列

14、a2nbn的前n項(xiàng)和為(3n-4)2n+2+16.方法總結(jié)方法總結(jié)(1)等差數(shù)列與等比數(shù)列中分別有五個(gè)量,a1,n,d(或q),an,Sn,一般可以“知三求二”,通過(guò)列方程(組)求關(guān)鍵量a1和d(或q),問(wèn)題可迎刃而解.(2)數(shù)列anbn,其中an是公差為d的等差數(shù)列,bn是公比q1的等比數(shù)列,求anbn的前n項(xiàng)和應(yīng)采用錯(cuò)位相減法.5.(2016四川,19,12分)已知數(shù)列an的首項(xiàng)為1,Sn為數(shù)列an的前n項(xiàng)和,Sn+1=qSn+1,其中q0,nN*.(1)若a2,a3,a2+a3成等差數(shù)列,求數(shù)列an的通項(xiàng)公式;(2)設(shè)雙曲線x2-=1的離心率為en,且e2=2,求+.22nya21e22

15、e2ne解析解析(1)由已知,Sn+1=qSn+1,Sn+2=qSn+1+1,兩式相減得到an+2=qan+1,n1.又由S2=qS1+1得到a2=qa1,故an+1=qan對(duì)所有n1都成立.所以,數(shù)列an是首項(xiàng)為1,公比為q的等比數(shù)列.從而an=qn-1.由a2,a3,a2+a3成等差數(shù)列,可得2a3=a2+a2+a3,所以a3=2a2,故q=2.所以an=2n-1(nN*).(2)由(1)可知,an=qn-1.所以雙曲線x2-=1的離心率en=.由e2=2解得q=.所以,+=(1+1)+(1+q2)+1+q2(n-1)=n+1+q2+q2(n-1)22nya21na2(1)1nq21q32

16、1e22e2ne=n+=n+(3n-1).2211nqq12易錯(cuò)警示易錯(cuò)警示(2)中求等比數(shù)列前n項(xiàng)和時(shí)要注意公比是q2.C C組組 教師專用題組教師專用題組考點(diǎn)一數(shù)列求和考點(diǎn)一數(shù)列求和1.(2012課標(biāo)全國(guó),12,5分)數(shù)列an滿足an+1+(-1)nan=2n-1,則an的前60項(xiàng)和為()A.3690B.3660C.1845D.1830答案答案D當(dāng)n=2k,kN*時(shí),a2k+1+a2k=4k-1,當(dāng)n=2k-1,kN*時(shí),a2k-a2k-1=4k-3,a2k+1+a2k-1=2,a2k+1+a2k+3=2,a2k-1=a2k+3,a1=a5=a61.a1+a2+a3+a60=(a2+a3)

17、+(a4+a5)+(a60+a61)=3+7+11+(260-1)=3061=1830.30 (3 119)22.(2019浙江,20,15分)設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,a3=4,a4=S3.數(shù)列bn滿足:對(duì)每個(gè)nN*,Sn+bn,Sn+1+bn,Sn+2+bn成等比數(shù)列.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)記cn=,nN*,證明:c1+c2+cn2,nN*.2nnabn解析解析本題主要考查等差數(shù)列、等比數(shù)列、數(shù)列求和、數(shù)學(xué)歸納法等基礎(chǔ)知識(shí),同時(shí)考查運(yùn)算求解能力和綜合應(yīng)用能力.體現(xiàn)了數(shù)學(xué)抽象、數(shù)學(xué)運(yùn)算等核心素養(yǎng).滿分15分.(1)設(shè)數(shù)列an的公差為d,由題意得a1+2d=4,a1+3

18、d=3a1+3d,解得a1=0,d=2.從而an=2n-2,nN*.所以Sn=n2-n,nN*.由Sn+bn,Sn+1+bn,Sn+2+bn成等比數(shù)列得(Sn+1+bn)2=(Sn+bn)(Sn+2+bn).解得bn=(-SnSn+2).所以bn=n2+n,nN*.1d21nS(2)cn=,nN*.我們用數(shù)學(xué)歸納法證明.當(dāng)n=1時(shí),c1=02,不等式成立;假設(shè)n=k(kN*)時(shí)不等式成立,即c1+c2+ck2,2nnab222 (1)nn n1(1)nn nk那么,當(dāng)n=k+1時(shí),c1+c2+ck+ck+12+2+2+=2+2(-)=2,即當(dāng)n=k+1時(shí)不等式也成立.根據(jù)和,不等式c1+c2+

19、cn2對(duì)任意nN*成立.k(1)(2)kkkk11k k21kk k1k k1k n思路分析思路分析(1)利用等比中項(xiàng)定義求出bn.(2)寫出cn,利用數(shù)學(xué)歸納法結(jié)合不等式放縮證明.一題多解(2)cn=,nN*.我們用數(shù)學(xué)歸納法證明.當(dāng)n=1時(shí),c1=02,不等式成立;假設(shè)n=k(kN*)時(shí)不等式成立,即c1+c2+c3+ck2,那么,當(dāng)n=k+1時(shí),只需證明c1+c2+ck+ck+12,即證2+2,即證k(k+1),所以,又,2nnab222 (1)nn n1(1)nn nk1k k(1)(2)kkk1k (1)(2)kkk1k k21kk (1)(2)kkk11k 11k 21kk 所以.

20、即當(dāng)n=k+1時(shí),不等式也成立.根據(jù)和,不等式c1+c2+cn0.因?yàn)閏kbkck+1,所以qk-1kqk,其中k=1,2,3,m.當(dāng)k=1時(shí),有q1;當(dāng)k=2,3,m時(shí),有l(wèi)nq.設(shè)f(x)=(x1),則f(x)=.令f(x)=0,得x=e.列表如下:lnkkln1kk lnxx21 lnxxx(1,e)e(e,+)f(x)+0-f(x)極大值因?yàn)?1時(shí),記cn=,求數(shù)列cn的前n項(xiàng)和Tn.nnab解析解析(1)由題意有,即解得或故或(2)由d1,知an=2n-1,bn=2n-1,故cn=,于是Tn=1+,Tn=+.-可得Tn=2+-=3-,故Tn=6-.111045100,2,ada d1

21、12920,2,ada d11,2,ad19,2.9ad121,2,nnnanb11(279),929.9nnnanb1212nn322523724921212nn1212232352472592212nn1212212212n212nn232nn1232nn7.(2015天津,18,13分)已知an是各項(xiàng)均為正數(shù)的等比數(shù)列,bn是等差數(shù)列,且a1=b1=1,b2+b3=2a3,a5-3b2=7.(1)求an和bn的通項(xiàng)公式;(2)設(shè)cn=anbn,nN*,求數(shù)列cn的前n項(xiàng)和.解析解析(1)設(shè)數(shù)列an的公比為q,數(shù)列bn的公差為d,由題意知q0.由已知,有消去d,整理得q4-2q2-8=0.

22、又因?yàn)閝0,解得q=2,所以d=2.所以數(shù)列an的通項(xiàng)公式為an=2n-1,nN*;數(shù)列bn的通項(xiàng)公式為bn=2n-1,nN*.(2)由(1)有cn=(2n-1)2n-1,設(shè)cn的前n項(xiàng)和為Sn,則Sn=120+321+522+(2n-3)2n-2+(2n-1)2n-1,2Sn=121+322+523+(2n-3)2n-1+(2n-1)2n,上述兩式相減,得-Sn=1+22+23+2n-(2n-1)2n=2n+1-3-(2n-1)2n=-(2n-3)2n-3,所以,Sn=(2n-3)2n+3,nN*.24232,310,qdqd評(píng)析評(píng)析本小題主要考查等差數(shù)列、等比數(shù)列及其前n項(xiàng)和公式等基礎(chǔ)知識(shí)

23、.考查數(shù)列求和的基本方法和運(yùn)算求解能力.8.(2015山東,19,12分)已知數(shù)列an是首項(xiàng)為正數(shù)的等差數(shù)列,數(shù)列的前n項(xiàng)和為.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=(an+1),求數(shù)列bn的前n項(xiàng)和Tn.11nnaa21nn2na解析解析(1)設(shè)數(shù)列an的公差為d.令n=1,得=,所以a1a2=3.令n=2,得+=,所以a2a3=15.解得a1=1,d=2,所以an=2n-1.(2)由(1)知bn=2n22n-1=n4n,所以Tn=141+242+n4n,所以4Tn=142+243+n4n+1,兩式相減,得-3Tn=41+42+4n-n4n+1=-n4n+1=4n+1-.121a a13

24、121a a231a a254(14 )14n1 33n43所以Tn=4n+1+=.319n4914(31)49nn9.(2014課標(biāo),17,12分)已知an是遞增的等差數(shù)列,a2,a4是方程x2-5x+6=0的根.(1)求an的通項(xiàng)公式;(2)求數(shù)列的前n項(xiàng)和.2nna解析解析(1)方程x2-5x+6=0的兩根為2,3,由題意得a2=2,a4=3.設(shè)數(shù)列an的公差為d,則a4-a2=2d,故d=,從而a1=.所以an的通項(xiàng)公式為an=n+1.(2)設(shè)的前n項(xiàng)和為Sn,由(1)知=,則Sn=+,Sn=+.兩式相減得Sn=+-=+-.所以Sn=2-.1232122nna2nna122nn2323

25、4212nn122nn12332442112nn222nn1234311122n222nn34141112n222nn142nn思路分析思路分析(1)解出方程的根,根據(jù)數(shù)列是遞增的得出a2,a4的值,從而解出通項(xiàng);(2)用錯(cuò)位相減法求和.10.(2013課標(biāo),17,12分)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.解析解析(1)設(shè)an的公差為d.由題意,得=a1a13,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

26、+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.從而Sn=(a1+a3n-2)=(-6n+56)=-3n2+28n.211a2n2n思路分析思路分析(1)設(shè)an的公差為d.由已知條件可建立方程求得公差d,從而求出an的通項(xiàng)公式.(2)由(1)得a3n-2是首項(xiàng)為25,公差為-6的等差數(shù)列,利用求和公式及等差數(shù)列的性質(zhì)可得答案.考點(diǎn)二數(shù)列的綜合應(yīng)用考點(diǎn)二數(shù)列的綜合應(yīng)用1.(2018江蘇,14,5分)已知集合A=x|x=2n-1,nN*,B=x|x=2n,nN*.將AB的所有元素從小到大依次排列構(gòu)成一個(gè)數(shù)列an

27、.記Sn為數(shù)列an的前n項(xiàng)和,則使得Sn12an+1成立的n的最小值為.答案答案27解析解析本題考查數(shù)列的插項(xiàng)問(wèn)題.設(shè)An=2n-1,Bn=2n,nN*,當(dāng)AkBlAk+1(k,lN*)時(shí),2k-12l2k+1,有k-2l-1k+,則k=2l-1,設(shè)Tl=A1+A2+B1+B2+Bl,則共有k+l=2l-1+l個(gè)數(shù),即Tl=,而A1+A2+=2l-1=22l-2,B1+B2+Bl=2l+1-2.則Tl=22l-2+2l+1-2,則l,Tl,n,an+1的對(duì)應(yīng)關(guān)系為121212lA12llS12lA2 1 1212l 2(12 )12llTlnan+112an+11323362104560330

28、79108494121720453182133396611503865780觀察到l=5時(shí),Tl=S2112a39,則n22,38),nN*時(shí),存在n,使Sn12an+1,此時(shí)T5=A1+A2+A16+B1+B2+B3+B4+B5,則當(dāng)n22,38),nN*時(shí),Sn=T5+=n2-10n+87.an+1=An+1-5=An-4,12an+1=122(n-4)-1=24n-108,Sn-12an+1=n2-34n+195=(n-17)2-94,則n27時(shí),Sn-12an+10,即nmin=27.22 55(22 1)()2nnAA2.(2017江蘇,19,16分)對(duì)于給定的正整數(shù)k,若數(shù)列an滿

29、足:an-k+an-k+1+an-1+an+1+an+k-1+an+k=2kan對(duì)任意正整數(shù)n(nk)總成立,則稱數(shù)列an是“P(k)數(shù)列”.(1)證明:等差數(shù)列an是“P(3)數(shù)列”;(2)若數(shù)列an既是“P(2)數(shù)列”,又是“P(3)數(shù)列”,證明:an是等差數(shù)列.證明證明本小題主要考查等差數(shù)列的定義、通項(xiàng)公式等基礎(chǔ)知識(shí),考查代數(shù)推理、轉(zhuǎn)化與化歸及綜合運(yùn)用數(shù)學(xué)知識(shí)探究與解決問(wèn)題的能力.(1)因?yàn)閍n是等差數(shù)列,設(shè)其公差為d,則an=a1+(n-1)d,從而,當(dāng)n4時(shí),an-k+an+k=a1+(n-k-1)d+a1+(n+k-1)d=2a1+2(n-1)d=2an,k=1,2,3,所以an-

30、3+an-2+an-1+an+1+an+2+an+3=6an,因此等差數(shù)列an是“P(3)數(shù)列”.(2)數(shù)列an既是“P(2)數(shù)列”,又是“P(3)數(shù)列”,因此,當(dāng)n3時(shí),an-2+an-1+an+1+an+2=4an,當(dāng)n4時(shí),an-3+an-2+an-1+an+1+an+2+an+3=6an.由知,an-3+an-2=4an-1-(an+an+1),an+2+an+3=4an+1-(an-1+an).將代入,得an-1+an+1=2an,其中n4,所以a3,a4,a5,是等差數(shù)列,設(shè)其公差為d .在中,取n=4,則a2+a3+a5+a6=4a4,所以a2=a3-d ,在中,取n=3,則a1

31、+a2+a4+a5=4a3,所以a1=a3-2d ,所以數(shù)列an是等差數(shù)列.方法總結(jié)方法總結(jié)數(shù)列新定義型創(chuàng)新題的一般解題思路:1.閱讀審清“新定義”;2.結(jié)合常規(guī)的等差數(shù)列、等比數(shù)列的相關(guān)知識(shí),化歸、轉(zhuǎn)化到“新定義”的相關(guān)知識(shí);3.利用“新定義”及常規(guī)的數(shù)列知識(shí),求解證明相關(guān)結(jié)論.3.(2015湖南,19,13分)設(shè)數(shù)列an的前n項(xiàng)和為Sn.已知a1=1,a2=2,且an+2=3Sn-Sn+1+3,nN*.(1)證明:an+2=3an;(2)求Sn.解析解析(1)證明:由條件,對(duì)任意nN*,有an+2=3Sn-Sn+1+3,因而對(duì)任意nN*,n2,有an+1=3Sn-1-Sn+3.兩式相減,得

32、an+2-an+1=3an-an+1,即an+2=3an,n2.又a1=1,a2=2,所以a3=3S1-S2+3=3a1-(a1+a2)+3=3a1.故對(duì)一切nN*,an+2=3an.(2)由(1)知,an0,所以=3.于是數(shù)列a2n-1是首項(xiàng)a1=1,公比為3的等比數(shù)列;數(shù)列a2n是首項(xiàng)a2=2,公比為3的等比數(shù)列.因此a2n-1=3n-1,a2n=23n-1.于是S2n=a1+a2+a2n=(a1+a3+a2n-1)+(a2+a4+a2n)=(1+3+3n-1)+2(1+3+3n-1)=3(1+3+3n-1)=,從而S2n-1=S2n-a2n=-23n-1=(53n-2-1).2nnaa3

33、(31)2n3(31)2n32綜上所述,Sn=3223(5 31),23(31),.2nnnn是奇數(shù)是偶數(shù)考點(diǎn)一數(shù)列求和考點(diǎn)一數(shù)列求和三年模擬A A組組 20172019 20172019年高考模擬年高考模擬考點(diǎn)基礎(chǔ)題組考點(diǎn)基礎(chǔ)題組1.(2019河北五個(gè)一名校聯(lián)盟第一次診斷,6)已知等差數(shù)列an中,a3+a5=a4+7,a10=19,則數(shù)列ancosn的前2018項(xiàng)的和為()A.1008B.1009C.2017D.2018答案答案D設(shè)an的公差為d,則有解得an=2n-1,設(shè)bn=ancosn,則b1+b2=a1cos+a2cos2=2,b3+b4=a3cos3+a4cos4=2,數(shù)列anco

34、sn的前2018項(xiàng)的和為(b1+b2)+(b3+b4)+(b2017+b2018)=2=2018.故選D.1112637,919,adadad11,2,ad20182解題關(guān)鍵解題關(guān)鍵由cosn的周期推得ancosn每?jī)身?xiàng)的和為2是解本題的關(guān)鍵.2.(2017湖南郴州第一次教學(xué)質(zhì)量監(jiān)測(cè),6)在等差數(shù)列an中,a4=5,a7=11.設(shè)bn=(-1)nan,則數(shù)列bn的前100項(xiàng)之和S100=()A.-200B.-100C.200D.100答案答案D設(shè)數(shù)列an的公差為d,由題意可得an=2n-3bn=(-1)n(2n-3)S100=(-a1+a2)+(-a3+a4)+(-a99+a100)=502=

35、100,故選D.1135,611adad11,2ad 3.(2019廣東珠海期末質(zhì)量監(jiān)測(cè),14)已知數(shù)列an的通項(xiàng)公式為an=2n+n,若數(shù)列an的前n項(xiàng)和為Sn,則S8=.答案答案546解析解析由an=2n+n,可得S8=(2+22+23+28)+(1+2+8)=+=546.82(12 )128 (1 8)24.(2018河北邯鄲第一次模擬,15)已知數(shù)列an,bn的前n項(xiàng)和分別為Sn,Tn,bn-an=2n+1,且Sn+Tn=2n+1+n2-2,則2Tn=.答案答案2n+2+n(n+1)-4解析解析由題意知Tn-Sn=b1-a1+b2-a2+bn-an=n+2n+1-2,又Sn+Tn=2n

36、+1+n2-2,所以2Tn=Tn-Sn+Sn+Tn=2n+2+n(n+1)-4.5.(2019湖南三湘名校(五十校)第一次聯(lián)考,15)已知數(shù)列an的前n項(xiàng)和為Sn,a1=1.當(dāng)n2時(shí),an+2Sn-1=n,則S2019=.答案答案1010解析解析由an+2Sn-1=n(n2),得an+1+2Sn=n+1,兩式作差可得an+1-an+2an=1(n2),即an+1+an=1(n2),所以S2019=1+1=1010.201826.(2018江西吉安一中、九江一中等八所重點(diǎn)中學(xué)4月聯(lián)考,13)若an,bn滿足anbn=1,an=n2+3n+2,則bn的前2018項(xiàng)和為.答案答案10092020解析

37、解析anbn=1,且an=n2+3n+2,bn=-,bn的前2018項(xiàng)和為-+-+-+-=-=.2132nn1(2)(1)nn11n12n121313141415120191202012120201010 12020100920207.(2018河南安陽(yáng)第二次模擬,17)設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,點(diǎn)(n,Sn)在函數(shù)f(x)=x2+Bx+C-1(B,CR)的圖象上,且a1=C.(1)求數(shù)列an的通項(xiàng)公式;(2)記bn=an(+1),求數(shù)列bn的前n項(xiàng)和Tn.12na解析解析(1)設(shè)數(shù)列an的公差為d,則Sn=na1+d=n2+n,又Sn=n2+Bn+C-1,兩式對(duì)照得解得所以a1=1,所

38、以數(shù)列an的通項(xiàng)公式為an=2n-1.(1)2n n2d12da1,210,dC 2,1,dC(2)由(1)知bn=(2n-1)(22n-1-1+1)=(2n-1)2n,則Tn=12+322+(2n-1)2n,2Tn=122+323+(2n-3)2n+(2n-1)2n+1,兩式相減得Tn=(2n-1)2n+1-2(22+2n)-2=(2n-1)2n+1-2-2212 (12)12n=(2n-3)2n+1+6.8.(2019廣東東莞第二次調(diào)研,17)已知數(shù)列an滿足a2=3,an+1=2an+1,設(shè)bn=an+1.(1)求a1,a3;(2)判斷數(shù)列bn是不是等比數(shù)列,并說(shuō)明理由;(3)求a1+a

39、3+a5+a2n+1.解析解析(1)當(dāng)n=1時(shí),a2=2a1+1,解得a1=1.當(dāng)n=2時(shí),a3=2a2+1,解得a3=7.(2)bn是等比數(shù)列.理由:因?yàn)閍n+1=2an+1,所以an+1+1=2(an+1),因?yàn)閎n=an+1,所以bn+1=2bn,所以bn是以2為公比的等比數(shù)列.(3)由(1)和(2)得an=2n-1,所以a1+a3+a5+a2n+1=(21+23+22n+1)-(n+1)=-(n+1)=-(n+1).12(14)14n12(41)3n考點(diǎn)二數(shù)列的綜合應(yīng)用考點(diǎn)二數(shù)列的綜合應(yīng)用1.(2019河北保定期末,10)在數(shù)列an中,若a1=1,a2=3,an+2=an+1-an(n

40、N*),則該數(shù)列的前100項(xiàng)之和是()A.18B.8C.5D.2答案答案Ca1=1,a2=3,an+2=an+1-an(nN*),a3=3-1=2,a4=2-3=-1,a5=-1-2=-3,a6=-3+1=-2,a7=-2+3=1,a8=1+2=3,a9=3-1=2,an是周期為6的周期數(shù)列,100=166+4,S100=16(1+3+2-1-3-2)+(1+3+2-1)=5.故選C.2.(2018福建漳州期末調(diào)研測(cè)試,5)等差數(shù)列an和等比數(shù)列bn的首項(xiàng)均為1,公差與公比均為3,則+=()A.64B.32C.38D.331ba2ba3ba答案答案D依題意知an=1+3(n-1)=3n-2,b

41、n=3n-1,則b1=1,b2=3,b3=9,+=a1+a3+a9=1+7+25=33.故選D.1ba2ba3ba3.(2017湖南湘潭三模,9)已知Tn為數(shù)列的前n項(xiàng)和,若mT10+1013恒成立,則整數(shù)m的最小值為()A.1026B.1025C.1024D.1023212nn答案答案C=1+,Tn=n+1-,T10+1013=11-+1013=1024-,又mT10+1013,整數(shù)m的最小值為1024.故選C.212nn12n12n101210124.(2018河北衡水中學(xué)八模,8)已知函數(shù)f(x)=ax+b(a0,且a1)的圖象經(jīng)過(guò)點(diǎn)P(1,3),Q(2,5).當(dāng)nN*時(shí),an=,記數(shù)列

42、an的前n項(xiàng)和為Sn,當(dāng)Sn=時(shí),n的值為()A.7B.6C.5D.4( )1( )(1)f nf nf n1033答案答案D函數(shù)f(x)=ax+b(a0,且a1)的圖象經(jīng)過(guò)點(diǎn)P(1,3),Q(2,5),或(舍去),f(x)=2x+1,an=-,Sn=+=-,令Sn=,得n=4.故選D.23,5,abab2,1ab1,4ab 121 1(21)(21)nnn 121n1121n113511591112121nn131121n1033特別提醒特別提醒使用裂項(xiàng)相消法求和時(shí),要注意正負(fù)項(xiàng)相消時(shí)消去了哪些項(xiàng),保留了哪些項(xiàng),切不可漏寫未被消去的項(xiàng),未被消去的項(xiàng)有前后對(duì)稱的特點(diǎn).5.(2019安徽黃山畢業(yè)

43、班第二次質(zhì)量檢測(cè),10)已知數(shù)列an和bn的前n項(xiàng)和分別為Sn和Tn,且an0,6Sn=+3an-4(nN*),bn=,若對(duì)任意的nN*,kTn恒成立,則k的最小值為()A.B.C.D.2na11(1)(1)nnaa1319112115答案答案B因?yàn)?Sn=+3an-4,所以6Sn+1=+3an+1-4,相減得6an+1=+3an+1-3an,因?yàn)閍n0,所以an+1-an=3,又6S1=+3a1-4,所以6a1=+3a1-4,因?yàn)閍10,所以a1=4,因此an=3n+1,bn=,從而Tn=,則Tn1,根據(jù)一次函數(shù)的單調(diào)性可得+10,由分段函數(shù)的單調(diào)性結(jié)合數(shù)列的單調(diào)性可得,2018-20200

44、,故可解得m2.故選C.2017,2019,312020,2019,2018nmnmnn32018m312018m7.(2017河北冀州第二次階段考試,15)若數(shù)列an是正項(xiàng)數(shù)列,且+=n2+3n,則+=.1a2ana12a23a1nan答案答案2n2+6n解析解析令n=1,得=4,所以a1=16.當(dāng)n2時(shí),+=(n-1)2+3(n-1).與已知等式聯(lián)立,化簡(jiǎn)得=(n2+3n)-(n-1)2-3(n-1)=2n+2(n2),所以an=4(n+1)2(n2),n=1時(shí),a1=16符合上式.所以an=4(n+1)2,nN*,所以=4n+4,+=2n2+6n.1a1a2a1nana1nan12a23

45、a1nan(844)2nn8.(2019湖南郴州第二次教學(xué)質(zhì)量監(jiān)測(cè),16)已知數(shù)列an和bn滿足a1a2a3an=(nN*),若數(shù)列an為等比數(shù)列,且a1=2,a4=16,則數(shù)列的前n項(xiàng)和Sn=.2nb1nb答案答案21nn解析解析an為等比數(shù)列,且a1=2,a4=16,公比q=2,an=2n,a1a2a3an=2122232n=21+2+3+n=.a1a2a3an=,bn=,=2,的前n項(xiàng)和Sn=b1+b2+b3+bn=2=2=.431aa3162(1)22n n2nb(1)2n n1nb2(1)n n111nn1nb111111111223341nn111n21nnB B組組 201720

46、1920172019年高考模擬年高考模擬專題綜合題組專題綜合題組(時(shí)間:65分鐘分值:80分)一、選擇題(每題5分,共20分)1.(2019河北石家莊模擬考試一(B卷),12)已知數(shù)列an的前n項(xiàng)和為Sn,且Sn+1+Sn=(nN*),若a10a11,則Sn取最小值時(shí)n的值為()A.10B.9C.11D.122192nn答案答案ASn+1+Sn=(nN*),Sn+Sn-1=(n2且nN*),兩式作差得an+1+an=n-10(n2且nN*),當(dāng)n=10時(shí),a11+a10=0,又a10a11,a100S10且S9S10,又S12-S10=a12+a11=11-10=10,由選項(xiàng)可得:Sn取最小值

47、時(shí)n的值為10,故選A.2192nn2(1)19(1)2nn思路分析思路分析將已知中的n換為n-1后,兩式作差得an+1+an=n-10(n2且nN*),分析可得a1000,可得結(jié)論.2.(2019河南許昌、洛陽(yáng)三模,11)已知數(shù)列an,bn的前n項(xiàng)和分別為Sn,Tn,且an0,6Sn=+3an,bn=,若kTn恒成立,則k的最小值為()A.B.C.49D.2na12(21)(21)nnnaaa171498441答案答案B當(dāng)n=1時(shí),6a1=+3a1,解得a1=3.當(dāng)n2時(shí),由6Sn=+3an,得6Sn-1=+3an-1,兩式相減并化簡(jiǎn)得(an+an-1)(an-an-1-3)=0,由于an0

48、,所以an-an-1-3=0,an-an-1=3,故an是首項(xiàng)為3,公差為3的等差數(shù)列,所以an=3n.則bn=,故Tn=b1+b2+bn=-,由于Tn是單調(diào)遞增數(shù)列,-,故k.故k的最小值為,故選B.21a2na21na12(21)(21)nnnaaa171118181nn1722311111118n17111781n149117(81)n149117(81)n1491491493.(2019福建質(zhì)量檢查測(cè)試,12)數(shù)列an中,a1=2,且an+an-1=+2(n2),則數(shù)列的前2019項(xiàng)的和為()A.B.C.D.1nnnaa21(1)na4036201920191

49、0104037201940392020答案答案Ban+an-1=+2(n2),-2(an-an-1)=n,整理得(an-1)2-(an-1-1)2=n,(an-1)2-(a1-1)2=n+(n-1)+2,又a1=2,(an-1)2=(n2),當(dāng)n=1時(shí),也適合上式,=2(nN*).則數(shù)列的前2019項(xiàng)和為21-+-+-=2=.故選B.1nnnaa2na21na(1)2n n21(1)na 2(1)n n111nn21(1)na121213120191202011202020191010思路分析思路分析由an+an-1=+2(n2),可得-2(an-an-1)=n,化為(an-1)2-(an-1

50、-1)2=n,利用“累加法”可得(an-1)2=,再利用裂項(xiàng)相消法求和即可.1nnnaa2na21na(1)2n n4.(2019福建廈門第一中學(xué)3月模擬,11)已知數(shù)列an的前n項(xiàng)和為Sn,直線y=x-2與圓x2+y2=2an+2交于An,Bn(nN*)兩點(diǎn),且Sn=|AnBn|2.若a1+2a2+3a3+nan+2對(duì)任意nN*恒成立,則實(shí)數(shù)的取值范圍是()A.(0,+)B.C.0,+)D.2142na1,21,2答案答案B圓心(0,0)到直線y=x-2,即x-y-2=0的距離d=2,由d2+=r2,且Sn=|AnBn|2,得22+Sn=2an+2,4+Sn=2(Sn-Sn-1)+2,n2,

51、即Sn+2=2(Sn-1+2)且n2.Sn+2是以a1+2為首項(xiàng),2為公比的等比數(shù)列.由22+Sn=2an+2,取n=1,解得a1=2,Sn+2=(a1+2)2n-1,則Sn=2n+1-2,an=Sn-Sn-1=2n+1-2-2n+2=2n(n2).22| 2 2 |221|2nnA B14a1=2適合上式,an=2n,nN*.設(shè)Tn=a1+2a2+3a3+nan=2+222+323+(n-1)2n-1+n2n,則2Tn=22+223+324+(n-1)2n+n2n+1,-Tn=21+22+23+2n-n2n+1=-n2n+1=2n+1-2-n2n+1=(1-n)2n+1+2.Tn=(n-1)

52、2n+1+2.a1+2a2+3a3+nan+2對(duì)任意nN*恒成立,(n-1)2n+1+2對(duì)任意nN*恒成立.設(shè)bn=,因?yàn)閎n+1-bn=-=,所以b1b4bnbn+1,故bn的最大值為b2=b3=.2(12 )12n2na112nn112nn2nn112nn22nn12所以.故選B.12思路分析思路分析由已知得Sn+2是以a1+2為首項(xiàng),2為公比的等比數(shù)列.求出Sn,進(jìn)一步求得數(shù)列an的通項(xiàng),然后利用錯(cuò)位相減法求得a1+2a2+3a3+nan,代入a1+2a2+3a3+nan1024的最小n的值為.2na2na答案答案9解析解析當(dāng)n=1時(shí),a1=4,當(dāng)n2時(shí),an=Sn-Sn-1=2n+1-

53、2n=2n,所以an=所以bn=所以Tn=當(dāng)n=9時(shí),T9=210+910+2=11161024;當(dāng)n=8時(shí),T8=29+89+2=5861024的最小n的值為9.4,1,2 ,2.nnn8,1,22 ,2,nnnn18,1,2(1)2,2.nnn nn6.(2019安徽蚌埠第二次教學(xué)質(zhì)量檢查,16)數(shù)列an滿足a1=1,|an-an-1|=n2(nN*且n2).若數(shù)列a2n-1為遞增數(shù)列,數(shù)列a2n為遞減數(shù)列,且a1a2,則a99=.答案答案4950解析解析由于數(shù)列a2n-1為遞增數(shù)列,數(shù)列a2n為遞減數(shù)列,可求得a2-a1=-22,a3-a2=32,a4-a3=-42,a5-a4=52,a6-a5=-62,a98-a97=-982,a99-a98=992.故a99=(a99-a98)+(a98-a97)+(a6-a5)+(a5-a4)+(a4-a3)+(a3-a2)+(a2-a1)+a1=992-982+972-962+52-42+32-22+1=99