2020高考數(shù)學(xué)培優(yōu)大一輪課件新題培優(yōu)練刷好題練能力(3)

1、高效演煉分層突破D.2.在數(shù)列an中,a=2, a?=2,叫一%= 1 + ( 1)n,nC N ,則a0的值為()A. 990B.1 000C. 1 100D.99解析：選 A.n 為奇數(shù)時(shí)，an+2an=0, an=2; n 為偶數(shù)時(shí)，an+2 an= 2, an=n.故 &。= 2X30+(2 + 4+ + 60)=990.3. (2019河北“五個(gè)一名校聯(lián)盟模擬)已知數(shù)列an滿足：an+1 = an an 1(n>2, nCN *), a1=1, a2=2, Sn 為數(shù)列an的前n項(xiàng)和,則S2 018=(A. 3B.C. 1D.解析：選 A.因?yàn)?an+ 1= an- a

2、n-1,a3 = 1, a4 = 1, a5 = - 2, a6基礎(chǔ)題組練1 .數(shù)列an的通項(xiàng)公式是an=(1)n(2n1),則該數(shù)列的前100項(xiàng)之和為()B. - 100C. 200D. 100解析：選 D.由題意知 S100=(1 +3)+(5+7)+( 197+ 199) = 2X 50= 100.故選=1, a7=1, a8 = 2,，故數(shù)列an是周期為6的周期數(shù)列，且每連續(xù) 6項(xiàng)的和為0,故S2 018 = 336x 0+ a2 017 + a2 018 = a + a2= 3.故選 A.今二十幣+為+ (n+1) 21的值為(A n+12 (n+2)B3_+1.4 2 (n + 2

3、)C.4-2 n+ 1n+ 2D" 一 " 十 二2 n+ 1 n+211解析：選C.因?yàn)樨耙?Jn (n+2)(n+1) - 1 n + 2n11 .1(n+ 1) 2 1一,111所以 2+ 2+ 2+ +_3 144. 2 n+ 1 n+2 ;5. (2019開(kāi)封調(diào)研)已知數(shù)列an滿足ai=1, an+i an=2n(n N*),則 向oi8等于()A. 22 018- 1B. 3 X 21 009 3C. 3X21 009 - 1D. 3 X 21 008 -2n+ 1解析： 選 B.a = 1, a2= 2,又史三=下-=2, a1an+ 1 anan +2所以旬

4、= ,所以丁 =2(-), Tn= -+ + - + =2(；-" + Snn (n+1) n n+1S1S2Sn1 2.所以a】' a?' a5,成等比數(shù)列；a2, a* a6,成等比數(shù)列，所以S2 018 =a1 + a2+ a3 + a4 + a5 + a6+ + a2 017 + 22 018 = (a+ a3+ a5 + + a2 017)+ (a2 + a4+ a6+ +1 0091 009、a2 018)=+=3 21 0093.故選 B.12126. (2019鄭州質(zhì)量預(yù)測(cè))已知數(shù)列an的前n項(xiàng)和為 = 1, a2=2,且an+22an+1+ an=

5、0(n e N),記 Tn=白+ (n C N ),則 T2 018 =.S1S2Sn解析：由an+22an +1 + an= 0(n C N ),可得an+2 + an= 2an+1,所以數(shù)列an為等差數(shù)列，公差 d=a2 a1 = 21 = 1,通項(xiàng)公式 an=a1+(n1)xd=1 + n 1 = n,則其前 n 項(xiàng)和 Sn =n a1 an)21一)=2(1-n+ 112n：)=一:n+1n+1故 丁2 018=2X2 018 2 018+ 1=4 036=2 019.n(n+ 1)，1211 .111_,1 1較案.4-036口水.2 01927.已知數(shù)列an中，a = 2,且=4(

6、an+1一 an)(nC N*),則其前 9 項(xiàng)和 S9= an解析：由已知，得 an + 1 = 4anan +1 4an,即 an+1 4anan+1 + 4an = (an+1 2an) =0,所以an+1= 2an,2X (1 29)dn所以數(shù)列an是首項(xiàng)為2,公比為2的等比數(shù)列，故 S9 =2 -21 2=1 022.答案：1 0228.已知數(shù)列解析：因?yàn)閍n滿足an+1 = 2+.ana2,且a = 2,則該數(shù)列的前 2 018項(xiàng)的和等于_1 p -1 :2所以a2= 1,-1.從而 a3= 2, a4 = 1,即得/,n=2k- 1 (kC N ),a1 = 2,又 an +1=

7、 2 + ya” an?J, n=2k ( kC N*),故數(shù)列的前2 018項(xiàng)的和等于S2 018=1009 X 1 +1 _3_027答案：等9. (2019唐山市莫擬)已知數(shù)列an滿足:*2+a1 a27' = -3(32n- 1), nCN*. an 8(1)求數(shù)列an的通項(xiàng)公式;an(2)設(shè) bn=lOg3,求+ +b1b2b2b3bnbn+1.1斛：了 1)=3,當(dāng)n>2時(shí)，因?yàn)閚 1一)an 1n 2 301)-3(32n 2-1)8當(dāng)n = 1, = 32n-1也成立, an所以 an= 2n 1.3(2)bn = log 3 = (2 n 1),1因?yàn)閎nbn+

8、 11_111(2n-1) (2n+1)2(2n-1 2n+ 1)? , 11所以 b+b+- +1bnbn+1111= 2(1 -3)+(3-1、, 15)+r(2n11 ) 2n 11= 2(i-i)2n+ 1n =.2n+ 1、，、一八、“，一，八 3 一、，八3an 110. (2019唐山市摸底考試)已知數(shù)列an的前n項(xiàng)和為Sn, Sn=-2- 求an；(2)若bn=(n- 1)an,且數(shù)列bn的前n項(xiàng)和為Tn,求Tn.解：(1)由已知可得，2s=3an1,所以 2Sn-1 = 3an-1 1(n>2),一得，2(SnSn 1)= 3an3an1,化簡(jiǎn)得an= 3an 1(n方

9、2),在中，令n=1可得，a1 = 1,所以數(shù)列an是以1為首項(xiàng)，3為公比的等比數(shù)列，從而有an= 3 一.(2)bn = (n-1)3n 1,Tn= 0X 30+ 1 X 31+ 2X 32+ (n- 1)X 3匚1,貝U 3Tn = 0X 31 + 1 X 32+ 2X 33+ + (n 1)X3n.一得，一2Tn= 31 + 32+ 33+ +3n 1-(n-1)X 3n33n=-(n- 1)x 3n1-3(32n) X3n3=2.(2n 3) X3n+3所以 Tn=.4綜合題組練1 .在數(shù)列an中，若an+1+(-1)nan=2n-1,則數(shù)列an的前12項(xiàng)和等于(A. 76B. 78C

10、. 80D. 82an+ 2 +解析：選 B.由已知 an+ + ( 1) an= 2n 1,得 an+2+( 1) + an+1= 2n + 1, an=(1)n(2n1)+(2n + 1),取 n=1, 5, 9 及 n=2, 6, 10,結(jié)果相加可得 Si2= ai+a2+a3 + a4+ aii + ai2= 78.故選 B.2,已知數(shù)列an的前n項(xiàng)和為S,ai=i,當(dāng)n> 2時(shí)，an+2Sn1= n，則。17的值為()A. 2 0i5B. 2 0i3C. i 008D. i 009解析：選 D.因?yàn)?an+ 2Sn i= n, n>2,所以 an+i +2Sn= n+ i

11、, n> i,兩式相減得 an+i+ an = i, n > 2.又 a=i,所以 S2 0i7=ai+(a2 + a3)+ (a2 0i6 + a2 0i7)= i 009,故選 D.3 .已知數(shù)列an,若an+i=an+an+2(nC N *),則稱數(shù)列an為“凸數(shù)列”.已知數(shù)列bn 為“凸數(shù)列"，且 bi=1, b2=- 2,則數(shù)列bn的前2 019項(xiàng)和為.解析：由“凸數(shù)列”的定義及 bi = 1, b2=2,得 b3= 3, b4= 1, b5=2, b6 = 3,b7=1, b8=-2,，所以數(shù)列bn是周期為6的周期數(shù)列，且 bi+b2+b3+b4+b5+b6=

12、0,于是數(shù)列bn的前2 019項(xiàng)和等于bi+b2+b3=4.答案：44 .(一題多解)(2019合肥卞II擬)數(shù)列an滿足：ai = l且an+i=(：)(n C N *),則33an 十 n數(shù)列an的前n項(xiàng)和Sn=.(n+1) an-i3an+ n 3解析：通解：an+1,兩邊同時(shí)取倒數(shù)得 -=3- +3an+nan + i(n+1) an n+1n,整理得 匚=0+3,所以 匚一n = 3,所以數(shù)列q是以'=3為首項(xiàng)，3為 (n+1) anan+1 anan+1 an" a1公差的等差數(shù)列，所以 7=3n,所以an = 1,所以數(shù)列an是常數(shù)列，所以Sn=n. an33優(yōu)

13、解：用歸納法求解， ai = 1,根據(jù)an+i = -n,可得 a2=1, a3=J, a4=1,所33an+ n3331 1 -一 n以猜想an=3,經(jīng)驗(yàn)證an + 1 = 3,從而Sn=3.5. (2019合肥卞II擬)已知等差數(shù)列an中，a5-a3=4,前n項(xiàng)和為a,且3, 21, S4 成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式；(2)令bn=(-1)n4,求數(shù)列bn的前n項(xiàng)和Tn.anan+1解：(1)設(shè)an的公差為 d,由 a5a3=4,得 2d = 4, d= 2.所以 S2=2a1+2, S3 1 = 3ai + 5, S4= 4ai + 12,又 S2, S31, S4成等比數(shù)列

14、，所以(3ai+5)2=(2ai + 2) (4ai + 12),解得a1=1,所以 an=2n 1.(2)bn = ( 1)n=( 1)n(+), anan + 12n 1 2n+1當(dāng)n為偶數(shù)時(shí)，Tn=(1+3)+(111113+5)-(5+7)+(二12n 11)+或12n+ 1)' .1所以 Tn=- 1+2n+ 12n2n+ 1當(dāng)n為奇數(shù)時(shí),111111Tn=-(1 + 3)+(3+5)-(5+7)+(-+ 2 n 31 ) ( 12n 12n 11十 ),2n+ 11 2n+ 2所以 Tn=-1 =.2n+ 12n+ 12n1r 一 ，2n+ 1所以Tn=2n+2n為偶數(shù) .n為奇數(shù)i 2n+16. (2019銀川質(zhì)檢)正項(xiàng)數(shù)列an的前n項(xiàng)和Sn滿足：S2(n2+n 1)&(n2+n) = 0. (1)求數(shù)列an的通項(xiàng)公式an；n +1、，一 一、_.一_, _* ，一,5(2)令bn=(n+2)2a2，數(shù)列bn的前n項(xiàng)和為Tn,證明：對(duì)于任息的nCN，都有<.解：(1)由 Sn (n + n 1)Sn(n +n)=0,得Sn(n2+n)(Sn+ 1) = 0.由于數(shù)列an是正項(xiàng)數(shù)列，所以Sn>0, Sn=n2+n.于是a1 = S1 = 2,當(dāng)n>