[2019高考数学二轮复习专题——数列课件及练习（两组试卷）]
2019高考数学二轮复习专题——数列课件及练习（两组试卷） 2019高考数学二轮复习专题——数列课件及练习（1） 等差数列、等比数列的基本问题 1.等差数列{an}前9项的和等于前4项的和,若a1=1,ak+a4=0,则k= . 2.已知在等比数列{an}中,a3=2,a4a6=16,则(a_7 "—" a_9)/(a_3 "—" a_5 )= . 3.(2018江苏南通中学高三考前冲刺练习)已知等差数列{an}的公差d=3,Sn是其前n项和,若a1,a2,a9成等比数列,则S5的值为 . 4.(2018南通高三第二次调研)设等比数列{an}的前n项和为Sn.若S3,S9,S6成等差数列,且a8=3,则a5= . 5.设数列{an}的首项a1=1,且满足a2n+1=2a2n—1与a2n=a2n—1+1,则数列{an}的前20项和为 . 6.(2018江苏锡常镇四市高三教学情况调研(二))已知公差为d的等差数列{an}的前n项和为Sn,若S_10/S_5 =4,则(4a_1)/d= . 7.已知Sn为数列{an}的前n项和,若a1=2,且〖〖S_n〗_+〗_1=2Sn,设bn=log2an,则1/(b_1 b_2 )+1/(b_2 b_3 )+…+1/(b_10 b_11 )的值是 . 8.(2018扬州高三第三次调研)已知实数a,b,c成等比数列,a+6,b+2,c+1成等差数列,则b的最大值为 . 9.(2018扬州高三第三次调研)已知数列{an}满足an+1+(—1)nan=(n+5)/2(n∈N*),数列{an}的前n项和为Sn. (1)求a1+a3的值; (2)若a1+a5=2a3. ①求证:数列{a2n}为等差数列; ②求满足S2p=4S2m(p,m∈N*)的所有数对(p,m). 10.(2018苏锡常镇四市高三教学情况调研(二))已知等差数列{an}的首项为1,公差为d,数列{bn}的前n项和为Sn,若对任意的n∈N*,6Sn=9bn—an—2恒成立. (1)如果数列{Sn}是等差数列,证明数列{bn}也是等差数列; (2)如果数列{b_n+1/2}为等比数列,求d的值; (3)如果d=3,数列{cn}的首项为1,cn=bn—bn—1(n≥2),证明数列{an}中存在无穷多项可表示为数列{cn}中的两项之和. 答案精解精析 1.答案 10 解析 S9=S4,则9a1+36d=4a1+6d,a1+6d=a7=0,则a4+a10=2a7=0,则k=10. 2.答案 4 解析 等比数列中奇数项符号相同,a3>0,则a5>0,又a4a6=〖a_5〗^2=16,则a5=4,从而a7=8,a9=16,则(a_7 "—" a_9)/(a_3 "—" a_5 )=("—" 8)/("—" 2)=4. 3.答案 65/2 解析 由题意可得a1a9=〖a_2〗^2,则由a1(a1+24)=(a1+3)2,解得a1=1/2,则S5=5×1/2+(5×4)/2×3=65/2. 4.答案 —6 解析 由S3,S9,S6成等差数列可得S3+S6=2S9,当等比数列{an}的公比q=1时不成立,则q≠1,(a_1 "(" 1"—" q^3 ")" )/(1"—" q)+(a_1 "(" 1"—" q^6 ")" )/(1"—" q)=2(a_1 "(" 1"—" q^9 ")" )/(1"—" q),化简得2q6—q3—1=0,q3=—1/2(舍去1),则a5=a_8/q^3 =—6. 5.答案 2056 解析 由题意可得奇数项构成等比数列,则a1+a3+…+a19=(1"—" 2^10)/(1"—" 2)=1023,偶数项a2+a4+…+a20=(a1+1)+(a3+1)+…+(a19+1)=1033,故数列{an}的前20项和为2056. 6.答案 2 解析 由〖S_1〗_0/S_5 =4得〖S_1〗_0=4S5,即10a1+45d=4(5a1+10d),则(4a_1)/d=2. 7.答案 19/10 解析 由〖〖S_n〗_+〗_1=2Sn,且S1=a1=2,得数列{Sn}是首项、公比都为2的等比数列,则Sn=2n.当n≥2时,an=Sn—Sn—1=2n—2n—1=2n—1,a1=2不适合,则an={■(2"," n=1"," @2^(n"—" 1) "," n≥2"," )┤故bn={■(1"," n=1"," @n"—" 1"," n≥2"," )┤所以1/(b_1 b_2 )+1/(b_2 b_3 )+…+1/(b_10 b_11 )=1+1/(1×2)+1/(2×3)+…+1/(9×10)=1+(1"—" 1/2)+(1/2 "—" 1/3)+…+(1/9 "—" 1/10)=2—1/10=19/10. 8.答案 3/4 解析 设等比数列a,b,c的公比为q(q≠0),则a=b/q,c=bq,又a+6=b/q+6,b+2,c+1=bq+1成等差数列,则(b/q+6)+(bq+1)=2(b+2),化简得b=3/2"—" (q+1/q) ,当b最大时q<0,此时q+1/q≤—2,b=3/2"—" (q+1/q) ≤3/4,当且仅当q=—1时取等号,故b的最大值为3/4. 9.解析 (1)由条件,得{■(a_2 "—" a_1=3", ①" @a_3+a_2=7/2 ",②" )┤ ②—①得a1+a3=1/2. (2)①证明:因为an+1+(—1)nan=(n+5)/2, 所以{■(a_2n "—" a_(2n"—" 1)=(2n+4)/2 ",③" @a_(2n+1)+a_2n=(2n+5)/2 ",④" )┤ ④—③得a2n—1+a2n+1=1/2. 于是1=1/2+1/2=(a1+a3)+(a3+a5)=4a3, 所以a3=1/4,又由(1)知a1+a3=1/2,则a1=1/4. 所以a2n—1—1/4=—(a_(2n"—" 3) "—" 1/4)=… =(—1)n—1(a_1 "—" 1/4)=0, 所以a2n—1=1/4,将其代入③式, 得a2n=n+9/4. 所以a2(n+1)—a2n=1(常数),所以数列{a2n}为等差数列. ②易知a1=a2n+1,所以S2n=a1+a2+…+a2n =(a2+a3)+(a4+a5)+…+(a2n+a2n+1) =n^2/2+3n. 由S2p=4S2m知p^2/2+3p=4(m^2/2+3m). 所以(2m+6)2=(p+3)2+27, 即(2m+p+9)(2m—p+3)=27,又p,m∈N*, 所以2m+p+9≥12且2m+p+9,2m—p+3均为正整数,所以{■(2m+p+9=27"," @2m"—" p+3=1"," )┤解得p=10,m=4, 所以所求数对为(10,4). 10.解析 (1)证明:设数列{Sn}的公差为d", ∵6Sn=9bn—an—2,① 6Sn—1=9bn—1—an—1—2(n≥2),② ①—②得6(Sn—Sn—1)=9(bn—bn—1)—(an—an—1),③ 即6d"=9(bn—bn—1)—d,所以bn—bn—1=(6d""" +d)/9(n≥2)为常数, 所以{bn}为等差数列. (2)由③得6bn=9bn—9bn—1—d,即3bn=9bn—1+d, 因为{b_n+1/2}为等比数列,所以(b_n+1/2)/(b_(n"—" 1)+1/2)=(3b_(n"—" 1)+d/3+1/2)/(b_(n"—" 1)+1/2)=(3(b_(n"—" 1)+1/2)+d/3 "—" 1)/(b_(n"—" 1)+1/2)=3+(d/3 "—" 1)/(b_(n"—" 1)+1/2)(n≥2)是与n无关的常数, 所以d/3—1=0或bn—1+1/2为常数. 当d/3—1=0时,d=3,符合题意; 当bn—1+1/2为常数时, 在6Sn=9bn—an—2中,令n=1,则6b1=9b1—a1—2,又a1=1,解得b1=1, 所以bn—1+1/2=b1+1/2=3/2(n≥2), 此时3+(d/3 "—" 1)/(b_(n"—" 1)+1/2)=3+(d/3 "—" 1)/(3/2)=1, 解得d=—6. 综上,d=3或d=—6. (3)证明:当d=3时,an=3n—2. 由(2)得数列{b_n+1/2}是以3/2为首项,3为公比的等比数列,所以bn+1/2=3/2×3n—1=1/2?3n.即bn=1/2(3n—1). 当n≥2时,cn=bn—bn—1=1/2(3n—1)—1/2(3n—1—1)=3n—1; 当n=1时,也满足上式, 所以cn=3n—1(n≥1). 设an=ci+cj(1≤i