【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

　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<j,i,j∈N*),则3n-2=3i-1+3j-1,即3n-(3i-1+3j-1)=2,

　如果i≥2,因为3n为3的倍数,3i-1+3j-1为3的倍数,所以3n-(3i-1+3j-1)为3的倍数,

　则2也为3的倍数,矛盾.

　所以i=1,则3n=3+3j-1,即n=1+3j-2(j=2,3,4,…).

　所以数列{an}中存在无穷多项可表示为数列{cn}中的两项之和.

　2019高考数学二轮复习专题--数列课件及练习（2）

　　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<j,i,j∈N*),则3n-2=3i-1+3j-1,即3n-(3i-1+3j-1)=2,

　如果i≥2,因为3n为3的倍数,3i-1+3j-1为3的倍数,所以3n-(3i-1+3j-1)为3的倍数,

　则2也为3的倍数,矛盾.

　所以i=1,则3n=3+3j-1,即n=1+3j-2(j=2,3,4,…).

　所以数列{an}中存在无穷多项可表示为数列{cn}中的两项之和.