高考数学三轮复习必做的数列综合题
Doasx
高考数学数列综合题
21.数列,,的各项均为正数,为其前项和,对于任意,总有aSa,,成等差数列. n,N*aSnnnnnn
(?)求数列,,的通项公式; an
nxln,,,(?)设数列,,的前项和为 ,且,求证:对任意实数(是常数,,2.71828)bb,T,,x,1,eneennn2an
和任意正整数,总有 2; ,Tnn
n,1*,,a,c,(n,N),,,,(?) 正数数列c中,.求数列c中的最大项. nn,1nn
2*2Saa,,(?)解:由已知:对于,总有 ?成立 n,Nnnn
22Saa,,? (n ? 2)? nn,,11n,1
222a,a,a,a,a?--?得 nnnn,n,11
,,,,?a,a,a,aa,a nn,1nn,1nn,1
a,aa,a,1?均为正数,? (n ? 2) nn,1nn,1
,,?数列a是公差为1的等差数列 n
22Saa,,又n=1时,, 解得a=1 1111
*?a,n.() n,Nn
n1xlnb,(?)证明:?对任意实数,和任意正整数n,总有?. ,x,1,en22nan
111111T,,,?,,1,,,?,? n222,,1,22,3n,1n12n
111111,1,1,,,,?,,,2,,2 223n,1nn
2a,c,2,c,2(?)解:由已知 , 211
3443a,c,3,c,3,a,c,4,c,4,2,322433 55a,c,5,c,5544
ccccc,,,,,... 易得 12234
Doasx
猜想 n?2 时,,,是递减数列. cn
1,x,lnxlnx1,lnxx,令 ,,,,fx,,则fx,,22xxx
,?当 ,,x,3时,lnx,1,则1,lnx,0,即fx,0.
?在内为单调递减函数. ,,,,3,,,fx
lnn,1,,n,1a,clnc,知由. nnn,1n,1
?n?2 时, ,,是递减数列.即,,是递减数列. lnccnn
3,,又cc, , ?数列c中的最大项为c,3. n122
f(0),12*n2(设f(x)=,定义f (x)= f,f(x),,a=(n?N). 1n+11nn f(0),21,xn
(1) 求数列,a,的通项公式; n
24n,n*(2) 若,Q=(n?N),试比较9T与 T,a,2a,3a,?,2nan2n2n1232n24n,4n,1Q的大小,并说明理由. n
22,11解:(1)?f(0)=2,a==,f(0)= f,f(0),=, 11n+11n1,f(0)42,2n
2,1f(0),11,f(0)f(0),1f1,(0)11n,1nnn?a==== -= -a. n+1n2f(0),24,2f(0)f(0),222n,1nn,2f1,(0)n
1111n,1?数列,a,是首项为,公比为-的等比数列,?a=(). ,nn4242(2)?T= a+2a+3a+…+(2n-1)a+2na, 2 n 1 2 3 2 n,1 2 n
111111?T= (-a)+(-)2a+(-)3a+…+(-)(2n-1)a+2na (,),,2 n1 2 32 n12 n222222
= a+2a+…+(2n,1)a,na. 2 32 n2 n
3两式相减,得T= a+a+a+…+a+na. 2 n12 32 n2 n2
Doasx
11,,2n1(),,,,4231111n11,,2n,12n2n,1?T=+n×(-)=-(-)+(-). 2n 124266242,12
3n,1111n112n2n,1T=-(-)+(-)=(1-). 2n 2n9926292
3n,1?9T=1-. 2n2n2
3n,1又Q=1-, n2(2n,1)
2 n2当n=1时,2= 4,(2n+1)=9,?9T,Q ; 2 nn
2 n2当n=2时,2=16,(2n+1)=25,?9T,Q; 2 nn
2nn2013n22当n?3时,, 2,[(1,1)],(C,C,C,?,C),(2n,1)nnnn
?9T,Q . 2 nn
x,0,
,3( 设不等式组所
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
示的平面区域为D,记D内的格点(格点即横坐标和纵坐标均为y,0nn,
,y,,nx,3n,
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)T, (3)记,若对于一切正整数n,总有T?m成立,求实数m的取值 nnn2
范围.
(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
,123n1n 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
Doasx
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,,2,12n
n,2当时n,,3,12n
?T
T>…>T 1234n
27 故T的最大值是T=T= n23227 ?m?。 2
na,11na,0a,1{}b4(已知,且,数列{}a的前项和为S,它满足条件.数列中,ba,?. lga,,1nnnnnnSan
{}b(1)求数列T的前项和; nnn
*bb,(2)若对一切都有,求的取值范围. anN,nn,1
nna,11aa(1),解:(1) ,? ,,1S,nSaa,1n
1aa(1),n,1当时,. aSa,,,11a,1
nn,1aaaa(1)(1),,n*naSS,,aanN,,()当?2时,=,? nn,,annn,1aa,,11
nnnnbann,lgaa,aalglga此时??=?, n
Doasx
23n?……=……+ Tbb,,,blg(23aaaa,,,na).n12n
23n设uaaa,,,,23……+, nan
naa(1),23,1nnn,1?(1),,,,,auaaa……, ana,,,nana,1nn,1naaa(1),? u,,.n2aa,,1(1)
nn,1naaa(1),?? ……6分 Ta,lg[].,n2aa,,1(1)
nn,1bbnaanaa,,,,lg(1)lg(2)由可得 nn,1
na,a,1?当时,由,可得 lg0a,n,1
nn**a,,,,1(),1,nNa ?对一切都成立, nN,n,1n,1
a,1?此时的解为.
n01,,annaa,,,(1),,?当时,由 可得 lg0a,n,11nn**0,,a??对一切都成立, (),01,nNa,,,nN,n,1n,12
1?此时的解为0,,a. 2
由?,?可知
1*bb,a,10,,a对一切,都有的的取值范围是或. ……14分 anN,nn,12
44(1)(1)xx,,,5、已知函数(x,0)。 fx(),44(1)(1)xx,,,
x,R(?)若且,则称为的实不动点,求的实不动点; fxx(),xfx()fx()
,(II)在数列中,,(),求数列的通项公式。 {}aa,2afa,(){}an,Nn1nn,1n42xx,,61解:(?)由及得 fx(),fxx(),344xx,
42xx,,6112422,,,,,,,xxxx32101或(舍去), x,,344xx,3,1x,1x,1x,,1所以或,即的实不动点为或; fx()
4444,,(1)(1)1(1)1aaaaa,,,,,,nnnnn,1a,,,,(II)由条件得,从而有 ,,n,1444(1)(1)1(1)1aaaaa,,,,,,nnnnn,1,,
Doasx
aa,,11nn,1ln4ln,, aa,,11nn,1
,,a,1a,11nlnln30,,由此及知:数列是首项为,公比为4的等比数列,故有 ln3ln,,a,1a,11n,,
n,14n,1aa,,1131,n,14,nn()。 ln4ln33,,,,,an,Nn,1n4aa,,1131,nn
16、已知函数,点,是函数图像上的两个点,且线段的,,f,,x,x,R,,fxPP,,,,Px,yPx,y22211112x4,2
1P中点的横坐标为( 2
P?求证:点的纵坐标是定值;
n,,,,,,?若数列的通项公式为,求数列的前m项的和; aa,f,,m,N,n,1,2,?,maS,,nmnnm,,
mm,1aa,m,N?若时,不等式恒成立,求实数的取值范围( aSSmm,1
1x,x,2,,1解:?由题可知:,所以, 122
xx1211444,,,,,,,,,,,,yyfxfx1212xxxx121242424242,,,,,,,, xxxx12124444441,,,,,,,x,xxxxx1212122,,,,424442444,,,,,
y,y112P 点的纵坐标是定值,问题得证( y,,P24
nm,n1,,,,f,f,m,n?由?可知:对任意自然数,恒成立( ,,,,mm2,,,,
m,m,m1221,,,,,,,,,,S,f,f,,f,f,f? 由于,故可考虑利用倒写求和的方法(即由于:,,,,,,,,,,mmmmmm,,,,,,,,,,
m,m,m1221,,,,,,,,,,S,f,f,?,f,f,f,,,,,,,,,,mmmmmm,,,,,,,,,, 1221mm,m,,,,,,,,,,,?,f,f,f,,f,f,,,,,,,,,,mmmmm,,,,,,,,,,,,,,,,1m,12m,2m,11m,,,,,,,,,,,,,,2S,f,f,f,f,,f,f,2f?,,,,,,,,,,,,,,,,,,,,mmmmmmmm,,,,,,,,,,,,,,,,,,,,所以, 所以,
11,,,m,1,2f(1),3m,1,,26
1,,S,3m,1 m12
11,,,,S,3m,1S,3m,2??, ? mm,11212
Doasx
mm,11aaa,,m, ?等价于 ? 12a,,0,,SS3m,13m,2,,mm,1
依题意,?式应对任意m,N恒成立(
1am,,0显然a,0,因为(),所以,需且只需对任意m,N恒成立(即:m,Na,03m,13m,2
3m,2a,对m,N恒成立( 3m,1
3m,53m,2,93m,2gm,,,记(m,N)(? , g,,,,m,1,gm,,,,03m,1,,,,3m,23m,13m,23m,1
55,,?()的最大值为g1,,? ( a,m,N,,gm22