6第六章 不等式、推理與證明

4n1n1n7n4n1n1n7n課后課作[組]111＋＋＋?>－為()A．7C．

B8D．10選

1＝－>，11[2016·]用數(shù)＋?＋1＋211>由k是()A．增加1B和兩k＋2k＋11C．增兩項(xiàng)k＋2k＋k＋D．nk1＋?k1kn，左?，“k”“nk1”

k1

n2n2n22323*22n2n2n22323*222222.k21安慶]用數(shù)學(xué)歸納法證明>n(≥5，n∈)＋證()A．n＝4C．＝6

Bn＝5D．n＝

取第又n5n．[2015·德]用2＋2?＋＝＋

3

－證為)A．1B1＋2C．＋2

D．1＋＋2

n1濰坊]某個命題與正整數(shù)有關(guān)(N)＝時該命題也成立，現(xiàn)已知n＝()A．當(dāng)n＝5B當(dāng)n＝C．當(dāng)n＝D．當(dāng)＝3時，該命題不成立D.

由nn成立逆否命題南昌]已知()＝1

＋

＋

3

?)

f(k＋與f(k是()A．f(＋1)＝f()(2k＋1)(2＋Bf1)＝f()(k＋1)C．f(1)＝f()(2k＋2)

2

2222222234nnnnn34nnnn2*,22232,2222222234nnnnn34nnnn2*,22232,32,3D．f(＋1)＝f()(2k＋1)A

f(k1)12＋3?)(2＋k2)f(kk＋1)(22)11設(shè)＋＋＋?S－＝nn＋1

n

?＋＋＋32＋11S1＋＋?n1

111＝＋?n1n223＋2列{}項(xiàng)和為S有(Snn1)算SS，，想S＝______.nn12

＋1

令n1,＝＝11

n＋＝212

n＋＋a＝3123想n＝n1[2015·]觀察下列式子1,1＋＋＋1,1＋＋3＋3＋1?n∈N

＋??＋＝________.n121323142?1??＝n.．云]觀察下列等式1＝＋2＝31＋2

32,2323233233233323323333332332232,23232332332333233233333323322n12n11a?1121a1212＝12＋4＝?個________．

1，得0)；第二個12

1＝(12)；第式1

1

＋

3(1＋3)；1＋4＋34)等式23?＝(123?

[組]n

湖北高]已知數(shù){}各項(xiàng)，bn(n∈)e為自n＋1(1)求函f)＝＋e并比較

ebbbbbb?(2)，11231n(3)令＝(a?數(shù)列{}{c}前n為n1nnn<en(1)(x∞f(x1.當(dāng)f(x)>0<0f(當(dāng)f(x)<0>0f(故f(x(－(0x>0f(x)<(0)0即x<e.1x得<e，

n

①(2)＝1

1b2＝＝2·2

2

(21)＝

2123123333aaa12n1k*k11212kk112122123123333aaa12n1k*k11212kk112123112121312n1×22×3×31n121n11n12nn；b1(3＝.1233b(1)a12n1時，bb當(dāng)n(kN且k≥1)(k1)a1k當(dāng)nk，b

(1)

1

，kbbbb(k1)kaaa12k12k

1

1

(

1

當(dāng)n1據(jù)n都的nn得T＋c＋＋＋n123n1(a)()(aaa)(aaa)11nbbbb1＋＋4n1bb＋＋＋×22××4111b＋

＋×

＋bn

<＋

12n12*12n12*3*1a＋<e＋ee＝ST<e.12nnn．江]＝{1,2,3}，＝{1,2,3，n

}(nN){(a，除b或b整除aX，b}(nn合Sn(1)寫出f(2)當(dāng)n≥出f(n(1)(6)當(dāng)n≥，f()＝6當(dāng)n6時，(6)6＋13k(N

k≥6)nk＋

kS(1k1)(2，k1)，1)中

333333k16t6(t5k12k1f(k1)f()2＋＋3(k＋1bk16t1則tkf(k1)(k1k＋(1)2＋ck16t則kt1k11k1f(k1)f()2＋＋2(k＋dk16t3則