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《步步高学案导学设计》20132014学年高中数学人教B版选修22精要课件第二章推理与证明章末复习课_图文

《步步高学案导学设计》20132014学年高中数学人教B版选修22精要课件第二章推理与证明章末复习课_图文







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2

















1 (1) 1,3,5,7,9...

{1}

{3,5}{7,9,11}

{13,15,17,19} ...





f(n)(nN*) n





_f(_n_)___n_3 _





11335823



79112733131517196443...

n f(n) n f(n)n3.

(2) RtABC ABcACb

BCa

a2b2c2

cos2Acos2B1



RtABC r

a2b2 2.











3

3 S1S2S3

S S21S22S23S2.





3



cos2cos2cos21.

3 abc

R

a2b2c2 2.

1 (1)________

________

AB P |PA||PB|2a>|AB|

P



a11an13an1 S1S2S3



an Sn



x2y21 Sr2 S



ab



(2) nN* sin xcos x1 sinnxcosnx

n1,2,3,4

n1 sin xcos x1.





n2 sin2xcos2x1.



n3 sin3xcos3x(sin2xcos2x)(sin xcos x)





sin xcos x(sin xcos x)

(sin xcos x)2(1)2 sin2x2sinxcos xcos2x1. sin xcos x0.



sin3xcos3x1(1)0(1)1.



n4 sin4xcos4x(sin3xcos3x)(sin xcos x)



sin xcos x(sin2xcos2x)(1)2011.





nN*sinnxcosnx(1)n.



























2
(0)2sin 21sincos . ()



2sin 21sincos



4sin cos 1sincos .



(0)sin >0.



4cos 1c1os .



41c1os 4(1cos )

1cos >0 1c1os 4(1cos )2

1 1cos

41cos

4

cos 12 3





41c1os 4(1cos )





2sin

21sicnos





()

1c1os 4(1cos )4 (1cos >0 cos 12 3)

4cos 1c1os .

(0)sin >0.



4sin cos 1sincos .





2sin 21sincos .

2

sins2in 2cos()ssiinn

.

sin(2)2cos()sin



sin[()]2cos()sin







sin()cos cos()sin 2cos()sin





sin()cos cos()sin

sin[()]sin

sin

sins2in 2cos()ssiinn

.













" p q"" p q"





" p



q"" p q"



3 f(x)ax2bxc(a0) ab c f(0)f(1) f(x)0

f(x)0 k





ak2bkc0





f(0)cf(1)abc ab





k k2n(nZ) ak2bk4n2a2nb



2n(2nab)

k k2n1(nZ) ak2bk(2n1)(2na

ab) f(x)0





" ......""



......"" ......""......"









3 ac2(bd) x2axb0 x2cxd0









1a24b<0 2c24d<0 a2c2<4(bd)





a2c22ac 4(bd)>2ac ac<2(bd)





























" nk1 "


""

4 nN*1n2(n1)3(n2) ...(n2)3(n1)2n116n(n1)(n2) (1) n1 116123

(2) nk





1k2(k1)3(k2)...(k2)3(k1)2k116k(k



1)(k2)


nk1 1(k1)2k3(k1)...(k1)3k2

(k1)1

1k2(k1)...(k1)2k1[123...k(k 1)]

16k(k1)(k2)12(k1)(k2)



16(k1)(k2)(k3)



nk1



nN*

4 {an} n Sn nN*(nSn) ybxr(b>0 b1b r )

(1) r



(2) b2 bn2(log2an1)(nN*)



nN*b1b1 1b2b2 1...bnbn 1> n1





(1) Snbnr

n2 Sn1bn1r.

anSnSn1bn1(b1)



b>0 b1



n2 {an} b



a1bra2b(b1)



aa21bbbbr1b r1.

(2) b2 (1) an2n1

bn2n(nN*)

22 144 1...2n2n 1> n1.



n1 32 2.



>





nk(kN*)



22 144 1...2k2k 1> k1

nk1

22 144 1...2k2k 122kk 13> k122kk 1322kk31.

nk1

22kk31 k2

2k2 3 k1k2



2k2 3k12 k2



k1k2





22kk31 k2

nk1

nN*b1b1 1b2b2 1...bnbn 1> n1


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