_2019高考数学二轮复习专题--数列课件及练习（与）2019高考数学二轮复习专题-函数与导数提分训练

　 2019高考数学二轮复习专题--数列课件及练习（与）2019高考数学二轮复习专题-函数与导数提分训练

　2019高考数学二轮复习专题--数列课件及练习

　　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-函数与导数提分训练

　函数的图象与性质

　1.设集合A=[-1,0],B={y|y=(1/2)^(x^2 "-" 1) "," x"∈" R},则A∪B=　　　　.

　 2.函数f(x)=ln(1-√(3"-" x))的定义域为　　　　.

　 3.已知函数f(x)是定义在R上的周期为2的奇函数,当0<x<1时,f(x)=8x,则f("-"

　19/3)的值为　　　　.

　 4.(2018南京师大附中高三年级模拟)“a=1”是“函数f(x)=(x+1)/x+sinx-a2为奇函数”的　　　　条件.(填“充分不必要”“必要不充分”“充要”或“既不充分也不必要”)

　 5.(2018江苏扬州中学高三年级第四次模拟)若函数f(x)=xln(x+√(a+x^2 ))为偶函数,则a=　　　　.

　 6.(2018江苏扬州高三第一次模拟)已知函数f(x)={■(log_(1/2) "(-" x+1")-" 1"," x"∈[-" 1"," k"]," @"-" 2"|" x"-" 1"|," x"∈(" k"," a"]," )┤若存在实数k使得该函数的值域为[-2,0],则实数a的取值范围是　　　　.

　 7.(2018江苏苏中地区四校高三联考)已知函数f(x)=x2-2|x|+4的定义域为[a,b],其中a<b,值域为[3a,3b],则满足条件的数组(a,b)为　　　　.

　 8.(2017无锡高三调研)已知函数f(x)={■((x^2+2x"-" 1)/x^2

　"," x≤"-"

　1/2 "," @log_(1/2)

　(1+x)/2 "," x>"-"

　1/2 "," )┤g(x)=-x2-2x-2.若存在a∈R,使得f(a)+g(b)=0,则实数b的取值范围是　　　　.

　 9.(2018江苏苏州中学高三检测)已知函数f(x)=a+1/(4^x+1)的图象过点(1",-"

　3/10).

　(1)判断函数f(x)的奇偶性,并说明理由;

　(2)若-1/6≤f(x)≤0,求实数x的取值范围.

　10.(2018江苏如东高级中学高三上学期期中)已知函数f(x)=ax2-2ax+2+b(a≠0)在区间[2,3]上有最大值5,最小值2.

　(1)求a,b的值;

　(2)若b<1,g(x)=f(x)-2mx在[2,4]上是单调函数,求实数m的取值范围.

　 答案精解精析

　1.答案　[-1,2]

　解析　因为x2-1≥-1,所以0<(1/2)^(x^2 "-" 1)≤2,则B=(0,2],所以A∪B=[-1,2].

　2.答案　(2,3]

　解析　要使函数f(x)=ln(1-√(3"-" x))有意义,则{■(1"-" √(3"-" x)>0"," @3"-" x≥0"," )┤解得2<x≤3,故该函数的定义域是(2,3].

　3.答案　-2

　解析　f("-"

　19/3)=f("-"

　19/3+6)=f("-"

　1/3)=-f(1/3)=-2.

　4.答案　充分不必要

　解析　若f(x)为奇函数,则f(-1)=-f(1),即sin(-1)-a2=-2-sin1+a2,a2=1,a=±1,故“a=1”是“函数f(x)=(x+1)/x+sinx-a2为奇函数”的充分不必要条件.

　5.答案　1

　解析　由题意知f(x)=f(-x),

　即xln(x+√(a+x^2 ))=-xln[(-x)+√(a+"(-" x")" ^2 )],

　所以xln(x+√(a+x^2 ))+xln(-x+√(a+x^2 ))=0,

　所以xln(x2+a-x2)=0,所以xlna=0,则a=1.

　6.答案　(1/2 "," 2]

　解析　作出函数f(x)的图象(图略),由图可得实数a的取值范围是(1/2 "," 2].

　7.答案　(1,4)

　解析　因为f(x)=(|x|-1)2+3≥3,所以3a≥3,a≥1,则函数f(x)=(x-1)2+3,x∈[a,b]单调递增,所以f(a)=3a,f(b)=3b,则a,b是方程f(x)=x2-2x+4=x的两根,且a<b,则a=1,b=4,故数组(a,b)为(1,4).

　8.答案　(-2,0)

　解析　当x≤-1/2时,令1/x=t,t∈[-2,0),则y=-t2+2t+1∈[-7,1),当x>-1/2时,(1+x)/2>1/4,

　f(x)=log_(1/2)

　(1+x)/2<2,所以f(x)的值域是(-∞,2).存在a∈R,使得f(a)+g(b)=0,则-g(b)=f(a)<2,即b2+2b+2<2,

　解得-2<b<0.

　9.解析　(1)f(x)是奇函数.理由:因为f(x)的图象过点(1",-"

　3/10),所以a+1/5=-3/10,解得a=-1/2,所以f(x)=1/(4^x+1)-1/2=(1"-" 4^x)/(2"(" 4^x+1")" ),f(x)的定义域为R.因为f(-x)=1/(4^("-" x)+1)-1/2=4^x/(4^x+1)-1/2=(4^x "-" 1)/(2"(" 4^x+1")" )=-f(x),所以f(x)是奇函数.

　(2)因为-1/6≤f(x)≤0,所以-1/6≤1/(4^x+1)-1/2≤0,

　所以1/3≤1/(4^x+1)≤1/2,

　所以2≤4x+1≤3,所以1≤4x≤2,解得0≤x≤1/2.

　10.解析　(1)f(x)=a(x-1)2+2+b-a.

　①当a>0时,f(x)在[2,3]上为增函数,

　故{■(f"(" 3")" =5"," @f"(" 2")" =2"," )┤所以{■(9a"-" 6a+2+b=5"," @4a"-" 4a+2+b=2"," )┤解得{■(a=1"," @b=0"." )┤

　②当a<0时,f(x)在[2,3]上为减函数,

　故{■(f"(" 3")" =2"," @f"(" 2")" =5"," )┤所以{■(9a"-" 6a+2+b=2"," @4a"-" 4a+2+b=5"," )┤解得{■(a="-" 1"," @b=3"." )┤

　故{■(a=1"," @b=0)┤或{■(a="-" 1"," @b=3"." )┤

　(2)因为b<1,所以a=1,b=0,即f(x)=x2-2x+2,

　g(x)=x2-2x+2-2mx=x2-(2+2m)x+2.

　若g(x)在[2,4]上单调,则(2+2^m)/2≤2或(2+2^m)/2≥4,

　所以2m≤2或2m≥6,即m≤1或m≥log26.

　故实数m的取值范围是(-∞,1]∪[log26,+∞).

　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.