高考数学数列经典题及答案解析
f(0),12n55(设,定义,其中n?N*. f(x),f(x),f[f(x)],a,1n,11nn1,xf(0),2n(1)求数列,a,的通项公式;(2)若, T,a,2a,3a,?,2na,n2n1232n
,56(设数列的前n项和为,点均在函数y,3x,2的图像上。 {}aS(,)()nSnN,nnn
3(?)求数列的通项公式;(?)设,是数列的前n项和,求使得{}b{}abT,nnnnaann,1
m,对所有都成立的最小正整数m。 nN,T,n20
57(在直角坐标平面上有一点列,对一切正整数,P(x,y),P(x,y)?,P(x,y)?n111222nnn
135y,3x,点位于函数的图象上,且的横坐标构成以为首项,,1为公差的等差PP,nn42数列. ,,xn
?求点的坐标;子?设抛物线列中的每一条的对称轴都垂直于轴,c,c,c,?,c,?Pxn123n
2第条抛物线的顶点为,且过点,记与抛物线相切于的直线的斜cPD(0,n,1)cDnnnnnn
111率为,求:. k,,?,nkkkkkk1223n,1n
258(已知抛物线,过原点作斜率1的直线交抛物线于第一象限内一点,又过点PPxy,411
11作斜率为的直线交抛物线于点,再过作斜率为的直线交抛物线于点,,如PPP32224
1此继续,一般地,过点作斜率为的直线交抛物线于点,设点( Pxy(,)PPnn,1nnnn2
(?)令,求证:数列是等比数列(并求数列的前项和为 bxx,,{}b{}bSnnnn2121,,nnn
159.已知数列,,中,在直线y=x上,其中n=1,2,3„. aanaa,,、点(、)2n11nn,2
(?)令 (?)求数列 ,,,,b,a,a,3,求证数列b是等比数列;a的通项;nn,1nnn
ST,,,,nn(?)设的前n项和,是否存在实数,使得数列,,,,bS、T分别为数列a、,,,nnnnn,,为等差数列,若存在,试求出.若不存在,则说明理由。 ,
1260(已知数列满足:. {a}a,2,a,2(1,)ann,n11n
(1)求数列{a}的通项公式; n
2n (2)设,试推断是否存在常数A,B,C,使b,(An,Bn,C),2n
,对一切都有a,b,b成立,说明你的理由; n,Nnn,1n
n,2 (3)求证: a,a,?,a,2,612n
0(x,0),61 f(x),,n[x,(n,1)],f(n,1)(n,1,x,n,n,N*),
(1)在[0,3]上作函数y=f(x)的图象
n1 (2)求证: 1,,2,fi()i,1
(3)设S(a) (a?0)是由x轴、y=f(x)的图象以及直线x=a所围成的图形面积,当n?N*
1时,试寻求与的关系 S(n),S(n,1)f(n,)2
62(已知定义在R上的单调函数,存在实数,使得对于任意实f(x)x0数总有恒成立. f(xx,xx),f(x),f(x),f(x)x,x010201212
(1)求x的值. 0
11a,,b,f(),1(2)若,且对任意正整数n,有,记f(x),1nn0nf(n)2
4S,aa,aa,?,aa,T,bb,bb,?,bb,比较与T的 Snn1223nn,1n1223nn,1n3
大小关系,并给出证明;
42(3)若不等式对任意a,a,?,a,[log(x,1),log(9x,1),1]n,n,n122113522不小于2的正整数n都成立,求x的取值范围.
63(在等差数列中,,,其中是数列的,,,,aS,,14S,a,,14Saan4455nn
22xy前项之和,曲线的方程是,直线的方程是。 y,x,3nC,,1lna4n
(1)求数列的通项公式; ,,an
(2)当直线与曲线相交于不同的两点,时,令ABClnnn
,求的最小值; ,,M,a,4,ABMnnnnn
(3)对于直线和直线外的一点P,用“上的点与点P距离的最小ll
值”定义点P到直线的距离与原有的点到直线距离的概念是等价的,l
若曲线与直线不相交,试以类似的方式给出一条曲线与直线CCllnn间“距离”的定义,并依照给出的定义,在中自行选定一个椭圆,Cn
求出该椭圆与直线的“距离”。 l
65(设向量, (n为正整数),函数在a,(x,2)b,(x,n,2x,1)y,a,b
[0,1]上的最小值与最大值的和为,又数列满足: b a,,nn
nn,,12999,,,,( nbnbbb,,,,,,,,,,,,,,,,121,,,,,,121nn,101010,,,,
(1) 求证:( a,n,1n
(2) (2)(求的
表
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达式( bn
(3) 若,试问数列中,是否存在正整数,使得对c a b,,, c k,,nnnn
于任意的正整数,都有成立,证明你的结论((注:nc c,nk
与表示意义相同) ,,a,(a,a)a,a,a1212
266(已知函数. fxxx,,,,
n11,(1)数列满足: ,,若对任意的aafa,a,0nN,,,,,,,n1nn,1a,12,1ii恒成立,试求的取值范围; a1
1(2)数列满足: ,,记,为数列bbfb,nN,b,1Sc,,,,,,,n1nn,1kn,b1n
nT7k的前项和, 为数列的前项积,求证. ccTkk,,,,,,nnk,ST10,1kkk
x,0,
,y,0设不等式组所表示的平面区域为D,记D内的格点67(nn,
,y,,nx,3n,
(格点即横坐标和纵坐标皆为整数的点)的个数为f(n)(n?N*).
(1)求f(1)、f(2)的值及f(n)的表达式;
n (2)设b=2f(n),S为{b}的前n项和,求S; nnnn
f(n)f(n,1) (3)记,若对于一切正整数n,总有T?m成立,T,nnn2
求实数m的取值范围.
,68(已知数列,,中,且点在直线上. a,1,x,y,1,0a,,Pa,a,,n,Nn1nn1,
(1)求数列的通项公式; ,,an
1111 (2)若函数求函数,,f(n),,,,?,n,N,且n,2,n,an,an,an,a123n
的最小值; f(n)
1 (3)设表示数列,,的前项和。试问:是否存在关于的bnb,,Snnnan
整式,使得 ,,gn
对于一切不小于2的自然数恒n,,,,S,S,S,?,S,S,1,gn123n,1n
成立,
若存在,写出的解析式,并加以证明;若不存在,试说,,gn
明理由(
69.一个三角形数表按如下方式构成:第一行依次写上n(n?4)个数,在上一行的每相邻两数的中间正下方写上这两数之和,得到下一行,依此类推(记数表中第i行的第j个数为f(i,j)(
(1)若数表中第i (1?i?n,3)行的数依次成等差数列,求证:
第i+1行的数也依次成等差数列;
(2)已知f(1,j)=4j,求f(i,1)关于i的表达式;
1
在(2)的条件下,若f(i,1)=(i+1)(a,1),b= ,试求一(3)iiaaii+1
n111个函数g(x),使得S=, ,m?( , ),均存在实数,使,bgi()n,i343,1i
得当n,时,都有 Sm,,n
n,1n70(已知数列中,,,,,,,,a,0q,0aa,a,q,qb,a,2n1nnn,1n
. n,1,2,3,?
,,an(I)求证数列是等差数列; ,,nq,,
2(II)试比较与的大小; bbb132
bbkn(III)求正整数,使得对于任意的正整数,?恒成立. nkbbk,1n,1
71(已知等差数列中,公差,其前项和为,且满足:S{a}nd,0nn
,( a,a,45a,a,142314
(?)求数列的通项公式; {a}n
Sn (?)通过公式b,构造一个新的数列(若也是等差{b}{b}nnnn,c
数列,求非零常数; c
b*n(?)求()的最大值( f(n),n,N(n,25),bn,1
2,R72(已知二次函数f (x)=x+ax()( a
16(1)若函数y = f (sinx +cosx) ()的最大值为,求,Rfx()3x3的最小值;
nn,13n,13n,,?,,(2)当a = 2时,设n?N*, S= , 求f(n)f(n,1)f(3n,1)f(3n)
3证:< S < 2 ; 4
(3)当a > 2时, 求证:
2222f (sinxlogsinx+cosxlogcosx) ?1 – a , 其中x?R, x , 22
,k,且x , k,(k?Z) ,2
73(已知公差大于零的等差数列的前n项和为S,且满足:{a}nn
a,a,117a,a,22,( 3425
(1)求数列的通项公式; {a}ann
Snb,(2)若数列是等差数列,且,求非零常数c; {b}nnn,c
64bn(3)若(2)中的的前n项和为,求证: {b}2T,3b,Tnnnn,1(n,9)bn,1
2112,55解:(1),2,,, f(0),f[f(0)],f(0)a,,11n,11n2241,f(0),n
2,1f(0),11,f(0)1,f(0)f(0),111n,1nnna,,,,,,,,a? nn,12f(0),24,2f(0)2f(0),22,1nnn,21,f(0)n
a11111n,1n,1?,?数列,a,上首项为,公比为的等比数列, a,(,),,,nn42a242n
(2) T,a,2a,3a,?,2na,2n1232n11111 ,T,(,)a,(,)2a,(,)3a,?,(,)2na,2n1232n22222
112n[1,()]311131n,2n,142T,,n,(,),两式相减得: (1)T,,2n2n2n1242921,2
S2n56解:(I)依题意得,即。 ,,32nn,,32,nSnn
22,,?2时,a; 当n(32)312(1)65nnnnn,,,,,,,,,,,,assnnn,1,,
2当n=1时,×-2×1-1-6×1-5 1,,3as11
所以。 ,,,5()nNa6nn
31111,,(II)由(I)得,因此,使得b,,,,n,,aannnn,,,,,(65)6(1)526561,,,,,1nn
m111m,,nN,,成立的m必须满足?,即m?10,故满足要求的最小整1,,,,,20202261n,,,
数m为10。
53x,,,(n,1),(,1),,n,57解:(1) n22
13535 ?,,,,,,?,,,,yxnPnn33,(,3)nnn4424
?(2)的对称轴垂直于轴,且顶点为.设的方程为:?cPcxnnn
23125n,n,2 (),y,ax,,24
222把代入上式,得,的方程为:。 D(0,n,1)?cy,x,(2n,3)x,n,1a,1nn
11111'?,,,, k,y|,2n,3()nx,0kkn,n,n,n,(21)(23)22123n,1n
1111111111 ,[(,),(,),?,(,)]?,,?,kkkkkk257792n,12n,31223n,1n
11111=(,),, 252n,3104n,6
点评:本例为数列与解析几何的综合题,难度较大。(1)、(2)两问运用几何知识算出. kn
221)因为、在抛物线上,故??,58解:(Pxy(,)Pxy(,)xy,4,xy,4nnnnnn,,,111nnnn,,11
yy,11nn,1又因为直线的斜率为,即,??代入可得PP,nn,1n2xx,2nn,1
22111xx,nn,1,,,,xxnn,1nn,24xx,22nn,1
111, 故?,,,,,,bxxxxxx()(),,,,nnnnnnn2121212221,,,,222322nnn,,,222
b11n,1是以 ,,{}bn44bn
4131为公比的等比数列;, SS,,,,,,(1)1nnnn3444
159解:(I)由已知得 aaan,,,,2,11nn,2
3313 aaa,,,,,,,,,11,2214424
又 baa,,,1,baa,,,1,nnn,1nnn,,,121
ananaa,,,,,(1)1nnnn,,11,baa,,11nnn,,11222 ?,,,,.baaaaaa,,,,,,1112nnnnnnn,,,,2111
31是以为首项,以为公比的等比数列. ?{}b,n24
3131n,1(II)由(I)知, b,,,,,,(),nn4222
3131 ?,,,,,aa1,?,,,,,aa1,nn,121n2222
3131,,,,,, ?,,,,,aaaa,,,,,1,1,32nn,12n,12222将以上各式相加得:
3111 ?,,,,,,,,,,,aan(1)(),n121n,2222
11,(1)n,13131322aannn ?,,,,,,,,,,,,,1(1)(1)2.n1nn,1122222,12
3 ?,,,an2.nn2
(III)解法一:
ST,,nn存在,使数列是等差数列. {},,2n
111 Saaann,,,,,,,,,,,,,,,,,,,,,,3()(12)2nn1212n22211(1),22nnn(1),1333nnnn,,22,,,,32n ,,,,,,,3(1)3.nn1222221,2
31,,(1)n313342Tbbb,,,,,,,,,,,,,,(1). 12nn,1nn12222,12
ST,,ST,,nnnn数列是等差数列的充要条件是、是常数 B){},(,,AnBAnn
2即 STAnBn,,,,,nn
223333nn,nn,31,又 ,,,,,,3(1)(1)ST,,,,,,,,3()nn,nnn12222222
ST,,,nn?当且仅当,即时,数列为等差数列. 10,,{},,22n
解法二:
ST,,nn存在,使数列是等差数列. {},,2n
nn(1),由(I)、(II)知, abn,,,22?,,,STn22nnn2
nn(1),,,,,22nTTnnn,,32,ST,,nn2 ,,T,n2nnn
31,,(1)n313342Tbbb,,,,,,,,,,,,,,(1)又 12nn,1nn12222,12ST,,3233n,,,nn ,,,,(),1nnn222
ST,,nn当且仅当时,数列是等差数列. ?{},,2n
aan,121n,n60解:(1)由已知 a,2,()a,即,2,1n,n22n(n,1)n
aa1n 是公比为2的等比数列,又 ?数列{},222n1
ann2n ?,2.?a,2,nn2n
2n (2) ?b,b,[An,(4A,B)n,2A,2B,C],2,n1n
22 若恒成立. a,b,b恒成立,则n,An,(4A,B)n,2A,2B,Cnn,n1
A,1A,1,,
,,?4A,B,0,B,,4 ,故存在常数A、B、C满足条件 ,,
,,2A,2B,C,0C,6,,
(3) a,a,?,a,(b,b),(b,b),?,(b,b),b,b12n2132n,1nn,11
2n,12n,1 ,[(n,1),4(n,1),6],2,6,(n,2n,3),2,6
2n,1n,1 ,[(n,1),2],2,6,2,6
61解:(1)当n=1即0
0 ? ,1n,1
n12n2n,2又 ? 1,,2,,2,fi()n,1n,1,1i
(3)由(1)图象中可知:S(n)―S(n―1)表示一个以f(n,1)、f(n)为底,n―(n―1)=1为
高的梯形面积(当n=1时表示三角形面积),根据(*)可得
211(1)(1)nn,nn,n S(n)―S(n―1)=[(1)()]1[] fn,,fn,,,,22222
211(1)nnn,n()[(1)](1)又可得 fn,,nn,,n,,fn,,,,22222
1 ?S(n)―S(n―1)= f(n,)2
62解:(1)令,得x,x,012
? f(0),f(x),2f(0),?f(x),,f(0).00
令,得 ? f(x),f(x),f(1),f(0),?f(1),,f(0).x,1,x,00012
由?,?得 为单调函数, ?x,1.f(x),f(1).?f(x)00(2)由(1)得, f(x,x),f(x),f(x),f(1),f(x),f(x),1121212
?f(n,1),f(n),f(1),1,f(n),2,f(1),1,
1, ?f(n),2n,1.(n,Z)?a,.n2n,1
1111又 ?f(1),f(,),f(),f(),f(1)2222
11 ?f(),0,b,f(),1.122
111111又, ?f(),f(,),f(),f(),f(1),2f(),1nn,1n,1n,1n,1n,1222222
111n,1 ?2b,2f(),2,f(),1,b.?b,().n,1nnn,1n222
1111111111 ?S,,,,?,(,,,,?,,)n1,33,5(2n,1)(2n,1)213352n,12n,1
11 ,(1,)22n,1
1111111110112n,1n32n,1 T,()(),()(),?,()(),,(),?,()n222222222
11n[1,()]21n24,,[1,()] 1341,4
42121211nn ?S,T,(1,),[1,()],[(),].nn332n,134342n,1nnnnnn,1,110, ?4,(3,1),C3,C3,?,C3,C,3n,1,2n,1nnnn
43114 ?S,T,(,),0.?S,T.nnnnn322n,134
(3)令, F(n),a,a,?,an,1n,22n
111FnFnaaa则 (,1),(),,,,,,,0.2n,12n,2n,1nnn4,14,32,1
12?当时, n,2,n,NF(n),F(n,1),?,F(2),a,a,.343512422 即 log(x,1),log(9x,1),2.?,[log(x,1),log(9x,1),1].111135352222,
,,1,0,x
,5112,,x,, 解得或 ,9,1,0,,x,1.x,933,,11x,,.249,1x,
63解:(1)?,?,又?,?,S,a,,14S,,14aS,,14a,1554444
4a,aa,a,,1441 ?,?,, S,,14a,,8,,,,2a,1d,,341132
?。 a,3n,11n
22,xy,,1,2a4,,,a,4x,6ax,5a,0 (2),由题意,知,nnnn
,y,x,3,
162,,n,,即,?或,即或,,16a,5a,0a,53n,11,53n,11,,5nnn3,即或时,直线与曲线相交于不同的两点。Cn,2n,6n,1ln
22165a,a,,525nn,, ,,M,a,4,AB4242,,,a,,,,,a,,,,nnnnnn424a,,,n
2,925,,,42,9n,,,n,6,,,,?时,的最小值为。 M87n,624,,,n,163,n,1,
(3)若曲线与直线不相交,曲线与直线间“距离”是:CCllnn曲线上的点到直线距离的最小值。 Cln
2 曲线与直线不相交时,,即,,,,,16a,5a,00,a,5Clnnnn
即,?, 3n,11,5n,3,4,5
?时,曲线为圆,?时,曲线为椭圆。 Cn,3,4Cn,55n
22xy 选,椭圆方程为,设椭圆上任一点,,1n,324
,它到直线的距离 M,,2cos,,2sin,l
2cos2sin3,,,,36,32,?椭圆到直线的距离为Cldd,,,,,3min3222
32,1032。 (椭圆到直线的距离为) C,3l422
n,4265解 (1)证:对称轴, 所以在[0,1]y,x,(n,4)x,2x,,,02
上为增函数 , a,(,2),(n,3),n,1n
n,1n,2999,,,,(2)由得 ,,nb,n,1b,?,2b,b,,,?,,1,,,,12n,1n101010,,,,
n,299,,,,,, = 两式相减得n,1b,n,2b,?,b,?,,1,,12n,11010,,
n,19,,b,b,?,b,,S,,12nn10,,
?当n,1时,b,S,111
n,219,, 当n,2时,b,S,S,,,,nnn,11010,,
,1???????当n1时,
,n,2,即b,19n,,,??当n,2时,,,1010,,,
,,2???????当n1时,
,n,2,,,cab(3)由(1)与(2)得 ,n,19nnn,,???当n,2时,,,1010,,,设存在自然数,使对,恒成立 c,ckn,Nnk
23当时, n,1c,c,,0,c,c212110
n,298,n,,当时,, 当时, ?c,cn,2n,8c,c,,,,n,1nn,1n10100,,
当时,,当时, c,cc,cn,8n,8n,1nn,1n所以存在正整数,使对任意正整数,均有 nk,9
c,c,?,c,c,c,c,?12891011
,解:(1)因为,所以. fxx,,21aa,,21,,nn,1于是,, 为等比数列, aa,,,121a,,10a,1,,,,nn,11n
n,1所以, aa,,,112,,n1
n,1111,,从而, ,,,aa112,,,,n1
n111111121,,有.故. a,3,,,,,,,,,1,1,,21n,1aaaa,11222112,,,,,,1i111n,12
(2)因为 , bfb,,,bb(1),,nn,1nn
bb1bb111nn12所以,,, bbb,,(1),,,c,,Tnnn,1nnbb11,,bbbbbb,,nn,12311nnnn
11111111,.即有.由S,,,,,,,c,,,1nkb,1bbbbbbbn,11211,,kkknn
112,显然,知,即.因为,bbb,,1bbb,,,1,2,3b,0bb,,,,kkk,1n123,kk12bbkk,1
1nn111117T116k,,,,,,,,所以 . ,,k,2,,2421666210,STb26,,11kk,1kkk1,667解(1) f(1)=3 f(2)=6
当x=1时,y=2n,可取格点2n个;当x=2时,y=n,可取格点n个
?f(n)=3n
n (2)由题意知:b=3n?2 n
123n,1n S=3?2+6?2+9?2+„+3(n,1)?2+3n?2 n
23nn+1 ?2S=3?2+6?2+„+3(n,1)?2+3n?2n
123nn+1?,S=3?2+3?2+3?2+„3?2,3n?2 n
2nn+1 =3(2+2+„+2),3n?2
n,12,2n,1 =3? ,3n21,2
n+1n+1 =3(2,2),3n
n+1?,S=(3,3n)2,6 n
n+1S=6+(3n,3)2 n
f(n)f(n,1)3n(3n,3) (3) T,,nnn22
(33)(36)nn,,
n,1Tn,2n,12,,3(33)nn,Tn2nn2
n,2当时n,,1,1 2n
n,2当时n,,2,1
2n
n,2当时n,,3,1
2n
在直线上,即,且,68解:(1)由点P(a,a)x,y,1,0a,a,1a,1nn,1n,1n1数列{}是以1为首项,1为公差的等差数列 an
,同样满足,所以 a,1,(n,1),1,n(n,2)a,1a,nn1n
111f(n),,,?,(2) n,1n,22n
11111f(n,1),,,?,, n,2n,3n,42n,12n,2
111111 f(n,1),f(n),,,,,,02n,12n,2n,12n,22n,2n,1
7f(2), 所以是单调递增,故的最小值是 f(n)f(n)12
11111S(3)b,可得, S,S,(n,2),,1,,,?,nnnn,1nnn23
, nS,(n,1)S,S,1nn,1n,1
(n,1)S,(n,2)S,S,1n,1n,2n,2
„„
S,S,S,1211
nS,S,S,S,S,?,S,n,1n1123n,1
,n?2 S,S,S,?,S,nS,n,n(S,1)g(n),n123n,1nn
故存在关于n的整式g(x)=n,使得对于一切不小于2的自然数n
恒成立
69解:(1)数表中第行的数依次所组成数列的通项为,则fij,1,i,1,,
由题意可得
fijfijfijfijfijfij,,,,,,,,,,,1,11,,1,2,(,1),,,,,,,,,,,,,,,,,,
(其中为第行数所组成的数列的公差) ,,,fijfij,2,,2ddi,,,,
(2) fjj1,4,,,
第一行的数依次成等差数列,由(1)知,第2行的数也依次成等?
差数列,依次类推,可知数表中任一行的数(不少于3个)都依次
成等差数列. 设第行的数公差为,则, dd,2diiii,1
iii,,,111 ,所以则dd,,,,,2422i1
ii,1i,, ,,,,222,122fififififi,11,11,221,12,,,,,,,,,,,,,,,,,,,
2iii,1ii,1 ,,,,22,122fi,,,,,,,,21,112fi,,,,,2412i,,,,,,,,
iii,1 ,,,,,,,21212ii,,,,
fi,1,,i(3)由,可得 fiia,111,,,a,,,,121,,,,,,iii,1
11111,,i所以= 令,则gi,2,,b,,,i,,iii,1ii,1aa22121,,,,2121,,,,,,,1ii
11111,所以 bgi,,S,,,,,inii,1n,1,3,,3212121
111113,m要使得,即,只要=, Sm,,,m,,mnn,1n,1,,3213213
1113,,n,1,,所以只要, ,,m,,?,,,013m21,,m,34413,,
33,,,,即只要,所以可以令, ,,,n,,,log11log1122,,,,,13,m13m,,,,
x则当时,都有.所以适合题设的一个函数为 Sm,gx,2n,,,,n
n,170解:(I)?,, q,0a,a,q,qn,1n
n,1aaqqaa,,,,an,1nn1n?,又,0,即数列是以0为首项,1,,,1,,n,1n,1nnqqqqq,,
为公差的等差数列
ann且,() ,,n,,,,1,2,3a,n,1q,n,1nnq
nnn(II) ,,b,a,2,n,1q,2nn
23?,, b,2b,2q,8b,q,4132
2223423? ,,,,,,b,bb,q,4,22q,8,q,8q,16,4q,16213
2432222 ,,,,,q,4q,8q,q,,q,4q,8,qq,2,4,0
2 ?b,bb213
71解:(?)? 数列,,是等差数列, an
? (又, a,a,a,a,14aa,45231423
,5,9aa,,22 ? ,或( ,,a,9a,533,,
? 公差,? ,( a,5a,9d,023
? ,( d,a,a,4a,a,d,13212
? ( a,a,(n,1)d,4n,3n1
12(?)? , S,na,n(n,1)d,n,2n(n,1),2n,nn12
2Sn,n2n ? b,,( ? 数列是等差数列, ,,bnnn,cn,c
? ( 2b,b,bn,1nn,2
2222(n,1),(n,1)2n,n2(n,2),(n,2)2,,, ? ( (n,1),cn,c(n,2),c
212n,n去分母,比较系数,得 ( ? ( c,,b,,2nn12n,2
2nn11f(n),,,(?) ?( 225(n,25),2(n,1)36n,26n,25n,,26n
251当且仅当n,,即时,取得最大值( f(n)n,5n36
,72解:?令t = sinx +cosx=2sin(x + ), 33?,?–2?t?2, ,Rx
2a22ay = t + at = (t + )– , 24
162当a < 0时, t =–2时,y= 4–2a = , 解得:a = , ,最大33
1112此时, f(x) = (x –)– ,?f(x) = –. 最小值399
162当a ?0时, t =2时,y= 4+2a = , 解得:a = . 最大33
1112此时, f(x) = (x +)– ,?f(x) = –. 最小值399
1综合上述,条件满足时, 的最小值为–. fx()9
nn,13n,13n,,?,,? ? S= f(n)f(n,1)f(3n,1)f(3n)
1111,,?,,=, n,2n,33n,13n,2
1111,,?,,; 设S(n ) =n,2n,33n,13n,2
1111,?,,则S(n+1 ) = n,3n,43n,43n,5
S(n+1 ) –S(n ) =
1311111,,,,>=>0 ; (3n,5)n,2(3n,5)(n,2)3n,33n,43n,5n,2
47453*,,?S(n )在时单调递增,?S = S(n )?S(1) = n,N606041111又,,?,,, n,2n,33n,13n,2
13?S < . (2n,1),2,,2n,2n,2
3?综上有: < S < 2成立. 4
,22(3) )?x?R, x , k,且x , k,(k?Z), ?sinx, cosx ?(0,1), ,2
2222又sinx+cosx =1, 故设t = sinx, 则有cosx= 1 – t ,
设f (t) = t logt + (1 – t ) log (1 – t ) (其中t?(0,1)) 22
tf′(t ) = logt + loge –log (1 – t ) – loge = log. 222221,t
11令f′(t ) = 0, 得t =,当0 < t < 时, f′(t ) < 0, 所以22
11f (t )在(0, )单调递减,当 < t <1时, f′(t ) > 0, 所以f (t )22
111在(,1)单调递增,?t = 时f (t)取最小值等于f() = 222
11111log+log= log= – 1. 22222222
2222即有sinsinx+cosx logcosx ?– 1 . x log22
a2当> 2时, f(x) = x+ax的对称轴x= < – 1, a ,2?f (x)在(– 1,+,)上单调递增,
2222?f(sinx logsinx+cosx logcosx) ?f (–1 ) = 1 – a . 22
73解:(1)为等差数列,?,又, a,a,a,a,22{a}a,a,117n342534
2? ,是方程的两个根 aan,22x,117,034
又公差,?,?, a,aa,13a,9d,03434
,2,9ad,1a,,11? ? ? a,4n,3,,n,3,13add,41,,
n(n,1)2(2)由(1)知, S,n,1,,4,2n,nn2
2Sn,n61512nb,,b,??,,?是等差数b,b,{b}n123n1,c2,c3,cn,cn,c
12列,?,??(舍去) 2b,b,b2c,c,0c,,c,02132
(3)由(2)得:
22n,n22 ,时2T,3b,2(n,n),3(2n,2),2(n,1),4,4n,1b,,2nnn,n11n,2
bnn6464,26464n取等号,时取等,,,,4n,329nbnnnn(,9)(,9),2(,1),10,9n,1n,,10n
64bn号,(1)、(2)式中等号不可能同时取到,所以2T,3b, nn,1(n,9)bn,1