最新高中数学必修5数列解答题及答案优秀名师资料Word格式文档下载.docx
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Sa,0,d,0,m1a,0m,1,
n,21(数列{a}的前n项和记为S,已知a,1,a,S(n,1,2,3„)(,nn1n1nn
Sn求证:
数列{}是等比数列(n
n,2证明:
?
a,S,S,a,S,,,,n1n1nn1nn
2SSn,1n?
(n,2)S,n(S,S),整理得nS,2(n,1)S,所以,(,,nn1nn1nn,1n
Sn故{}是以2为公比的等比数列(n
aSaaSn,1,211,,,,2、数列的前项和记为n,,,,nnnn11,
a(?
)求的通项公式;
,n
b(?
)等差数列的各项为正,其前项和为,且,又T,15ababab,,,,,Tn,,nn3112233成等比数列,求Tn
aSn,,,212解:
(?
)由可得,aS,,21,,nn,1nn,1
aaaaan,,,,2,32两式相减得,,nnnnn,,11
n,1a13又?
故是首项为,公比为得等比数列?
aS,,,213aa,3a,3,,n2121n
bd(?
)设的公差为由得,可得,可得T,15bbb,,,15b,5,,n31232故可设又aaa,,,1,3,9bdbd,,,,5,513123
2515953,,,,,,dd由题意可得解得dd,,2,10,,,,,,12
nn,1,,2b?
等差数列的各项为正,?
?
d,0d,2Tnnn,,,,,322,,nn2
23、已知数列{a}的前n项和S,3n,2n,求证数列{a}成等差数列.nnn
111b,cc,aa,b
(2)已知,,成等差数列,求证,,也成等差数列.abcabc证明:
(1)n,1时,a,3,2,1,,S11
22当n?
2时,a,S,S,3n,2n,[3(n,1),2(n,1)],6n,5,,nnn1
n,1时,亦满足,?
a,6n,5(n?
N*)(n
a,6n,5,[6(n,1),5],6(常数)(n?
N*),首项a,1,a,1nn1
数列{a}成等差数列且a,1,公差为6(n1
111211
(2)?
,,成等差数列,?
,化简得2ac,b(a,c)(abcbac
222222,bacacbc,c,a,abac(ac)b,ca,b(,),,(,),,,,,,b(ac)acacacac
2
a,cb,cc,aa,b,2?
,?
,,也成等差数列(babc
4、设{a}是公比为q的等比数列,且a,a,a成等差数列(n132
(1)求q的值;
(2)设{b}是以2为首项,q为公差的等差数列,其前n项和为S,当n?
2时,比较Snnn
与b的大小,并说明理由(n
22解:
(1)由题设2a,a,a,即2aq,a,aq,?
a?
0,?
2q,q,1,0,3121111
1?
q,1或,(2
2,3n(n,1)nn
(2)若q,1,则S,2n,,(n22
(n,1)(n,2)当n?
2时,Sb,S,,0,故Sb(,,,nnn1nn2
21n(n,1)1,,9nn若q,,,则S,2n,(,),(n2224
(n,1)(10,n)当n?
2时,S,b,S,,,nnn14
故对于n?
N,当2?
n?
9时,S,b;
当n,10时,S,b;
当n?
11时,S,b(+nnnnnn
n,25、数列{a}的前n项和记为S,已知a,1,a,S(n,1,2,3„)(,nn1n1nn
2SSn,1nn,2)S,n(S,S),整理得nS,2(n,1)S,所以,(?
(,,nn1nn1nn,1n
26((理)(2011?
四川广元诊断)已知数列{a}的前n项和S,2n,2n,数列{b}的前n项和Tnnnn,3,b.n
求数列{a}和{b}的通项公式;
nn
11?
设c,a?
b,求数列{c}的前n项和R的表达式(nnnnn43
[解析]?
由题意得a,S,S,4n,4(n?
2)而n,1时a,S,0也符合上式,nnn111
b1n?
a,4n,4(n?
N)又?
b,T,T,b,b~?
,,,nnnn1n1n2b,n1
11133,n1n,,,,?
{b}是公比为的等比数列~而b,T,3,b~?
b,~?
b,,3?
(n?
N)(,n1111n,2,,2,222
111111nn,,,,?
C,a?
b,(4n,4)×
×
3,(n,1)~nnn,2,,2,4343
1111234n,,,,,,,,?
R,C,C,C,„,C,,2?
,3?
,„,(n,1)?
n123n,2,,2,,2,,2,
11111,34nn1,,,,,,,,?
R,,2?
,„,(n,2),(n,1)n,2,,2,,2,,2,2
111111,23nn1n,,,,,,,,,,?
R,,,„,,(n,1)?
~?
R,1,(n,1).nn,2,,2,,2,,2,,2,2
7((理)(2011?
华安、连城、永安、漳平、龙海、泉港六校联考)已知数列{b}前n项和为S,nn
1且b,1,b,S.,1n1n3
(1)求b,b,b的值;
(2)求{b}的通项公式;
(3)求b,b,b,„,b的值(234n2462n
1111141116[解析]
(1)b,S,b,~b,S,(b,b),~b,S,(b,b,b),.2113212431233333393327
1b,S?
,n1n,314
(2),b,b~?
b,b~?
?
解b,,n1nnn1n,331b,S?
nn1,3
1,n,1,,,411,n2,,?
b,~?
b,?
(n?
2)?
b,,.412nn,n2,,,3,33?
,n?
2,,,3,,3
412,,(3)b~b~b„b是首项为~公比的等比数列~2462n,3,3
142n][1,,,33342n?
b,b,b,„,b,,[(),1](2462n4732,,1,,3,
28((理)(2011?
黑龙江林口四中)已知a,2,点(a,a)在函数f(x),x,2x的图象上,其中,1nn1
n,1,2,3,„.
(1)证明数列{lg(1,a)}是等比数列;
n
(2)设T,(1,a)(1,a)„(1,a),求T及数列{a}的通项(n12nnn
22[解析]
(1)由已知a,a,2a~?
a,1,(a,1).?
a,2~?
a,1>
1~两边取对数得:
,,n1nnn1n1n
,lg,1,a,n1lg(1,a),2lg(1,a)~即)}是公比为2的等比数列(,2.?
{lg(1,a,n1nnlg,1,a,n
,,n1n1n1
(2)由
(1)知lg(1,a),2?
lg(1,a),2?
lg3,lg32n1
n1?
1,a,32(*)n
,01n12n1n?
T,(1,a)(1,a)„(1,a),32?
32?
„?
32,31,2,2,„,2,32,1.n12n
n由(*)式得a,32,1,1.n
29((理)(2011?
安徽河历中学月考)设曲线y,x,x,2,lnx在x,1处的切线为l,数列{a}的n首项a,,m,(其中常数m为正奇数)且对任意n?
N,点(n,1,a,a,a)均在直线l,,1n1n1上(
(1)求出{a}的通项公式;
(2)令b,na(n?
N),当a?
a恒成立时,求出n的取值范围,使得b>
b成立(,,nnn5n1n
2[解析]
(1)由y,x,x,2,lnx~
1,,,2x,1,知x,1时~y,4~又y′|,,2~,,x1x1,x,,
直线l的方程为y,4,2(x,1)~即y,2x,2~又点(n,1~a,a,a)在l上~a,,,n1n11m~
a,a,m,2n.即a,a,2n,m(n?
N)~,,,n1nn1n
a,a,2,m~a,a,2×
2,m~„a,a,2×
(n,1),m.,2132nn1
2各项迭加得~a,2(1,2,„,n,1),(n,1)m,a,n,(m,1)n.n1
2?
通项a,n,(m,1)n(n?
N),n
m,1m,1
(2)?
m为奇数~?
为整数~由题意知~a是数列{a}中的最小项~?
5~5n22
32322?
m,9~令f(n),b,n,(m,1)n,n,10n~则f′(n),3n,20n~n
20由f′(n)>
0得~n>
N)~,3
20即n>
N)时~f(n)单调递增~即b>
b成立~?
n的取值范围是n?
7~且n?
N.,,,n1n3
10((理)(2011?
辽宁丹东四校协作体联考)数列{a}满足a,1,a,2,n12
nπnπ22,,1,cosa,a,sin,n,1,2,3,„.,n2n,2,2
(1)求a,a,并求数列{a}的通项公式;
34n
a,12n1
(2)设b,,S,b,b,„,b.证明:
6时,|S,2|<
.nn12nnan2n
ππ22[解析]
(1)因为a,1~a,2~所以a,(1,cos)a,sin,a,1,2~1231122
22a,(1,cosπ)a,sinπ,2a,4.422
,2k,1,π,2k,1,π*22当n,2k,1(k?
N)时~a,[1,cos]a,sin,a,1~,,,2k12k12k122即a,a,1.所以a,k.,,,2k12k12k1
2kπ2kπ2*2,,1,cos当n,2k(k?
N)时~a,a,sin,2a.,2k22k2k,2,2
n,1*n,2k,1,k?
N,,2k所以a,2.故数列{a}的通项公式为a,2knn,n*2n,2k,k?
N,,2
a,n123n1123n2n1
(2)由
(1)知~b,,~S,,,,„,~?
S,,,,„,~?
n23n234,n1nnna22222222222n
11n,,,,1,2,,,,211111nn1n?
得~S,,,,„,,,,1,,.,23nn,,,n1n1n1n22222212221,2
n,21n所以S,2,,,2,.nn,n1n222
n,n,2,1要证明当n?
6时~|S,2|<
成立~只需证明当n?
6时~<
1成立(nnn2
6×
,6,2,483
(1)当n,6时~,,<
1成立(62644
k,k,2,
(2)假设当n,k(k?
6)时不等式成立~即<
1.k2
,k,1,,k,3,k,k,2,,k,1,,k,3,,k,1,,k,3,则当n,k,1时~,×
<
1~k,k1222k,k,2,,k,2,?
2k
n,n,2,1由
(1)、
(2)所述可知~当n?
1.即当n?
成立(nn2n
11((理)(2011?
湖南长沙一中期末)已知数列{a}和等比数列{b}满足:
a,b,4,a,b,2,nn1122
*a,1,且数列{a,a}是等差数列,n?
N.,3n1n
(1)求数列{a}和{b}的通项公式;
1*,,,3
(2)是否存在k?
N,使得a,b?
若存在,求出k的值;
若不存在,请说明理由(kk,2,
11,,n1n3,,,,[解析]
(1)易知b,4?
~?
a,a,,2~a,a,,1~„n2132,2,,2,
a,a,,2,(n,1),n,3.?
a,a,(n,1),3~,,n1nnn1
n,n,1,?
a,(a,a),(a,a),„,(a,a),(a,a),a,,3(n,1),4,,,nnn1n1n2322112
2n,7n,14,.2
2,7n,141n,n3,,
(2)设c,a,b,,.显然~当n,1,2,3时~c,0.nnnn,2,2
22,2n,1,7n,7,14n,7n,14111n,,,n2n3n2,,,,,,由c,c,,,,,n,3,.,n1n,2,,2,,2,22
11当n,3时~c,c,~?
c,a,b,,4344422
157当n,4时~c,c,1,,~?
c,a,b,,54555444
117313,,当n,5时~c,c,2,,~?
c,a,b,>
3,65666,2,88
1,n2,,当n?
6时~c,c,n,3,>
3恒成立,,n1n,2,
此时c,a,b>
3,c>
3恒成立(,,,n1n1n1n
1,,~3?
存在k,5~使a,b?
.kk,2,
12、{a}是公差为1的等差数列,{b}是公比为2的等比数列,P,Q分别是{a},{b}的前nnnnnnn
项和,且a,b,P,Q,45.63104
(1)求{a}的通项公式;
(2)若P>
b,求n的取值范围(nn6
(1)由题意得
a,5,4b11,,a,3,1,4,,3,(n,1),n,2.,?
~?
a,1,2,10×
9bn1b,2,10a,,,45,11,21,2,
2,5nn,n,2,3,n,61
(2)P,,~b,2×
2,64.n622
2n,5n2由>
64?
n,5n,128>
0?
n(n,5)>
128~2
*又n?
N~n,9时~n(n,5),126~?
10时~P>
b.n6[解析]?
2)而n,1时a,S,0也符合上式?
a,4n,,nnn111n
b11n4(n?
N)又?
b,T,T,b,b~?
{b}是公比为的等比数列~而b,T,3,,,nnn1n1nn1122b,n1
1133,n1n,,,,,b~?
N)(,11n,2,,2,22
3,(n,1)~nnn,2,,2,4343
2(理)(2011?
b,求数列{c}的前n项和R的表达式(nnnnnnn43
18((本小题满分12分)(文)(2011?
河南濮阳)数列{a}的前n项和记为S,a,1,a,,nn1n12S,1(n?
1)(n
(2)等差数列{b}的各项为正数,前n项和为T,且T,nnn315,又a,b,a,b,a,b成等比数列,求T.112233n
1(理)(2011?
六校联考)已知数列{b}前n项和为S,且b,1,b,S.,nn1n1n3
(3)求b,b,b,…,b的值(234n2462n
19((本小题满分12分)(文)(2011?
宁夏银川一中模拟)在各项均为负数的数列{a}中,已n
28*知点(a,a)(n?
N)在函数y,x的图象上,且a?
a,.,nn125327
(1)求证:
数列{a}是等比数列,并求出其通项;
(2)若数列{b}的前n项和为S,且b,a,n,求S.nnnnn
黑龙江)已知a,2,点(a,a)在函数f(x),x,2x的图象上,其中n,,1nn1
1,2,3,….
(2)设T,(1,a)(1,a)…(1,a),求T及数列{a}的通项(n12nnn
n20((本小题满分12分)数列{b}的通项为b,na(a>
0),问{b}是否存在最大项,证明你nnn
的结论(
x21((本小题满分12分)(2011?
湖南长沙一中月考)已知f(x),m(m为常数,m>
0且m?
1)(设
2f(a),f(a),…,f(a)…(n?
N)是首项为m,公比为m的等比数列(12n
数列{a}是等差数列;
(2)若b,af(a),且数列{b}的前n项和为S,当m,2时,求S;
nnnnnn
(3)若c,f(a)lgf(a),问是否存在正实数m,使得数列{c}中每一项恒小于它后面的项,nnnn
若存在,求出m的取值范围;
若不存在,请说明理由(
22((本小题满分12分)(文)(2011?
四川资阳模拟)数列{a}的前n项和为S,且S,n(n,nnn
*1)(n?
N)(
(1)求数列{a}的通项公式;
bbbb123n
(2)若数列{b}满足:
a,}的通项公式;
,,,…,,求数列{b23nnnn3,13,13,13,1
bann*(3)令c,),求数列{c}的前n项和T.(n?
Nnnn4
(1)当n,1时~a,S,2~11
2时~a,S,S,n(n,1),(n,1)n,2n~知a,2满足该式,nnn11?
数列{a}的通项公式为a,2n.nn
bbbb123n
(2)a,,,,„,(n?
1)?
23nn3,13,13,13,1
bbbbb,n1123n?
a,,,,„,,?
,,23nn1n13,13,13,13,13,1
b,n1,n1?
得~,a,2~b,2(3,1)~,a,,,n1n1nn13,1
n故b,2(3,1)(n?
N)(n
bannnn(3)c,,1),n?
3,n~,n(3n4
23n?
T,c,c,c,„,c,(1×
3,2×
3,3×
3,„,n×
3),(1,2,„,n)n123n
23n令H,1×
3~?
,234n1则3H,1×
3?
n,3,1,3,,23nn1n1?
得~,2H,3,3,3,„,3,n×
3,,n×
3n1,3
,n1,3,2n,1,×
H,~n4
数列{c}的前n项和n
,n1,3n,n,1,,2n,1,×
3T,,.n42
11,,n1n3,,,,(理)[解析]易知b,4?
~n,2,,2,
a,a,,2~a,a,,1~„2132
a,a,,2,(n,1),n,3.,n1n
a,a,(n,1),3~,nn1
(理)(2011?
a,b,4,a,b,2,ann11223
*,1,且数列{a,a}是等差数列,n?
N.求数列{a}和{b}的通项公式;
,n1nnn
(1)由题意得
9bn1b,2,,,,4510a,11,21,2,
b.n6(理)[解析]?
2)而n,1时a,S,0也符合上式,nnn111
b11n?
{b}是公比为的等比数列~,,,nnnn1n1nn22b,n1
1133,n1n,,,,而b,T,3,b~?
N)(,1111n,2,,2,22
18、(文)[解析]
(1)由a,2S,1可得a,2S,1(n?
2)~,,n1nnn1
两式相减得a,a,2a~?
a,3a(n?
2)~又a,2S,1,2a,1,3~?
a,3a~,,nnn1n21121n1
n1故{a}是首项为1~公比为3的等比数列~?
a,3.nn
(2)设{b}的公差为d~由T,15得~b,b,b,15~可得b,5~故可设b,5,d~n312321
2b,5,d~又a,1~a,3~a,9~由题意可得(5,d,1)(5,d,9),(5,3)~解得d,2或3123
10.
n,n,1,2?
等差数列{b}的各项均为正数~?
d,2~b,3~?
T,3n,×
2,n,2n.n1n2
1111141116(理)[解析]
(1)b,S,b,~b,S,(b,b),~b,S,(b,b,b),.2113212431233333393327
,n1n,3141
(2),b,b~?
b,b~?
解b,,n1nnn1n2,3331b,S?
1,n,1,,,41,n2,