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高中数学竞赛数列讲义


2014?b??9

?p?ê?Jpù?

ù?< ?? 2014c1

8?
1 2 3 n ?ê ê?1 ê?2 2 14 16

1

1.

n



~K 1.1. ???sin 2o s

in 18o sin 22o sin 38o sin 42o sin 58o sin 62o sin 78o sin 82o y?. |^? ? 4 sin x sin(60 + x) sin(60 ? x) = sin 3x : (sin 2o sin 58o sin 62o )(sin 18o sin 42o sin 78o )(sin 22o sin 38o sin 82o ) = sin 6o sin 54o sin 64o 64 o sin 18 = 256 √ 5?1 = 1024 (1) (2) (3)

~K 1.2. ???tan2 5o + tan2 10o + ... + tan2 85o y?. |^? ? tan2 x + tan2 (60 + x) + tan2 (60 ? x) = 9 tan 3x + 6 ? (tan2 5+tan2 55+tan2 65)+(tan2 15+tan2 45+tan2 75)+...+(tan2 25+tan2 35+tan2 65)+tan2 30+tan2 60 = 9(tan2 15 + tan2 30 + ... + tan2 75) + 30 + tan2 30 + tan2 60 = 9(tan2 15 + tan2 45 + tan2 75) + 63 + = 9(9 tan2 45 + 6) + 63 + = ps:?? ·?k
n?1

(4) (5) (6) (7)

1 3

1 3

595 3
2n

tan2
k=1

kπ = 3 2n 2n

~K 1.3. z{? 1 + 4 sin2 a cos2 a + 4 sin a cos a+ cos4 a ? 5 sin4 a ? 4 sin a cos a + 6 sin2 a y?. = (1 + 2 sin a cos a)2 + cos4 a ? sin4 a ? 4 sin4 a ? 4 sin a cos a + 6 sin2 a (8) (9)

= 1 + 2 sin a cos a +

cos2 a ? sin2 a ? 4 sin4 a ? 4 sin a cos a + 6 sin2 a 2

= 1 + 2 sin a cos a + = 1 + 2 sin a cos a + =2

1 + 4 sin2 a cos2 a ? 4 sin a cos a (1 ? 2 sin a cos a)2

(10) (11) (12)

~K 1.4. e2 sin x sin y ? 3 cos x cos y = 0?? 1 1 + 2 2 2 sin x + 3 cos x 2 sin y + 3 cos2 y
2

y?. ^?? 2 tan x tan y = 3 1 1 sin2 x + cos2 x sin2 y + cos2 y + = + 2 sin2 x + 3 cos2 y 2 sin2 y + 3 cos2 y 2 sin2 x + 3 cos2 y 2 sin2 y + 3 cos2 y tan2 x + 1 tan2 y + 1 + 2 tan2 x + 3 2 tan2 y + 3 5 = 6 = a b c = = ?y? tan (x + y ) tan (x + z ) tan (y + z ) (13) (14) (15)

~K 1.5. ex, y, z, a, b, c÷v

a+b b+c c+a sin2 (y ? z ) + sin2 (z ? x) + sin2 (x ? y ) = 0 a?b b?c c?a y?. d^? : tan(x + y ) + tan(x + z ) sin(2x + y + z ) a+b = = a?b tan(x + y ) ? tan(x + z ) sin(y ? z ) ·?k: a+b cos(2x + 2z ) ? cos(2x + 2y ) sin2 (y ? z ) = sin(2x + y + z ) sin(y ? z ) = a?b 2 ?? T??0 ~K 1.6. z{?sin8 x + cos8 y + 6 sin4 x cos4 x + 4 sin2 x cos2 x(sin4 x + cos4 x) y?. 3(a + b)4 ? a = sin2 x, b = sin2 y ~K 1.7. 3n /ABC ?,?y? √ 1 + cos A 1 + cos B 1 + cos C 2 2 2 + + >3 3 A B C y?. 3? ?cos x 1? x2 A ? x= 2 2 1 + cos A 2 A ?? : LHS 2 2 2 π + + ? A B C 8 2 A ? A 8 √ 18 π ? >3 3 π 8 T??1

3

~K 1.8. 3b n /ABC ?,?y? √ √ π 3 3 4sin A+sin B +sin C + 2tan A+tan B +tan C > 21+ 2 y?. d??? ? : LHS |^? ?2 sin x + tan x 3x, 2 2
2(sin A+sin B +sin C ) B +tan C + tan A+tan 3 3

π >x 2

0

: 3π

2(sin A + sin B + sin C ) + (tan A + tan B + tan C ) ‘\ y ps: π a, b, c > 0, x ∈ (0, )y? 2 2 a2 + b2 + c2 sin x a + b + c tan x + >a+b+c 3 x 3 x

~K 1.9. 3n /ABC ?,e2A + 3B = π y?4(a + b) < 5c y?. ^?? A = π ? π B 3B , C = + d u?n??Iy?: 2 2 2
B sin 32 + sin B sin A + sin B 5 5 < ? < B sin C 4 4 cos 2

|^

ú?,·?Psin

B B π A π 1 = t 5? :0 < = ? < ?0 < t < e??Iy 2 2 6 3 6 2
B 2

B sin 32 + sin B

cos

= ?4t2 + 2t + 1 <

5 ? (4t ? 1)2 4

0

~K 1.10. 3n /ABC ?,?

? u45o ,?y: 4(2 ? √ 2)

cot A + cot B + cot C + 3 cot A cot B cot C y?. ?” A = min (A, B, C )Kk |^cot x + cot y = 1 ? cot x cot y cot B + cot C π 4 A π 3

cot B cot C = 1 ? cot A(cot B + cot C )

:

LHS = cot A+cot B +cot C +3 cot A(1?cot A(cot B +cot C )) = 4 cot A+(cot B +cot C )(1?3 cot2 A) u45o ?Tn /7,?b n / π 5? :(cot x) = 2 cot x csc2 x > 0, x ∈ (0, ) 2 B+C Kdjensen? ??k:cot B + cot C 2 cot 2 (?1 ? 3 cot2 A 0?e??Iy?? d? ? 4 cot A + 2 cot B+C (1 ? 3 cot2 A) 2 4 4(2 ? √ 2)

e?Ptan

A = tKdA 2

√ 1 ‰??t ∈ [ 2 ? 1, √ ] K? 3 LHS = 4 ? 3(1 ? t2 )2 2t 4(2 ? √ 2)

~K 1.11. 3n /ABC ?,?y? (sin A)sin B + (sin B )sin C + (sin C )sin A > 1.19 y?. d??|? ?: (sin A)sin B = 1 (1 +
1 sin A

?

1)sin B

1+

1 ( sin A

1 sin A > sin A + sin B ? 1)(sin B )

-tan

A B C = a, tan = b, tan = cKab + bc + ca = 1 K: 2 2 2 LHS sin A = sin A + sin B : ab + ac 1 + ab a(1 + b2 ) a(1 + b2 ) + b(1 + a2 )

2|^1 + b2 = ab + bc + ca + b2 = (b + a)(b + c) a(1 + b2 ) = a(1 + b2 ) + b(1 + a2 ) 2d…?? ?? ab + ac = 1 + ab ab + 1 + ab ac 1 + ab 1 1+

a(a + b)(b + c) = a(a + b)(b + c) + b(b + a)(a + c)

a2 b2

+

1 1+ a2 bc

=

1 1+ a2 b2

+

2 3? a2 b2

√ (1 + 2)2 > RHS 4

~K 1.12. 3n /ABC ?,?e??f ???(n? sin y?. d?z? , ? A 2
n

ê)

A 2

n

sin

B C sin 2 2

= sin ·?-

C B +C cos B ? 2 ? cos 2 2

sin

A 2

n

1 A (1 ? sin ) 2 2

sin A = aKd??? ? 2
n

1 n a (n ? na) 2n

n (n + 1)
n

2(n + 1)

5

~K 1.13. ?O ?n / /G??(?dn 1. cos A = sin B + sin C ? 3 2 3 2

/n? ?

2. sin(B ? A) sin C + sin A + cos B = 3. cos 2A + 4. √ 3(cos 2B + cos 2C ) +

5 =0 2

sin A sin C sin B = √ = 1 2 3 √ √ 5. cos 2A + 2 2 cos B + 2 2 cos C = 3 √ 3+ 3 6. sin A + sin B + sin C = 2 7. sin A B C B C A C A B 9 cos cos + sin cos cos + sin cos cos = 2 2 2 2 2 2 2 2 2 8 A B C 1 A B+C B?C cos cos = sin (cos + cos ) 2 2 2 2 2 2 2

y?. 5? ? sin = ?? : cos A + cos B + cos C = ù ?A = B = C ~K 1.14. ??tan 0 ;tan B B C A , tan ÷v?§x2 ?a1 x+b1 = 0;tan , tan ÷v?§x2 ?a2 x+b2 = 2 2 2 2 3 2

1 A 1 B+C B?C 1 ? cos A cos B + cos C sin2 + cos cos = + 2 2 2 2 2 4 4

C A , tan ÷v?§x2 ? a3 x + b3 = 0,?y 2 2 (1 ? a1 + b1 )(1 ? a2 + b2 )(1 ? a3 + b3 ) √ 5616 ? 3240 3

y?. 5? : (x2 + a1 x + b1 )(x2 + a2 x + b2 )(x2 + a3 x + b3 ) = (x ? tan -x = ?1,e??Iy: (1 + tan 5? :tan A 2 B C ) (1 + tan )2 (1 + tan )2 2 2 2 tan A + 4 tan √ 5616 ? 3240 3 A 2 B C ) (x ? tan )2 (x ? tan )2 2 2 2

π?A B+C = tan ? 4 4 (1 + tan

A B A B C tan = 1 + tan tan tan K: 4 4 4 4 4

A 2 B C A B C ) (1 + tan )2 (1 + tan )2 = 4(1 + tan tan tan )2 2 2 2 4 4 4

6

A B C π A B C du . , ∈ (0, ) tan tan tan ∈ (0, 1) 2 2 2 4 4 4 4 A B C 3 ·?Ptan tan tan = x : 4 4 4 1 + x3 = 1 + tan A B C A A B tan tan = tan + tan tan 3x + 3x2 4 4 4 4 4 4 √ √ √ ←→ (x + 1)(x ? 2 + 3)(x ? 2 ? 3) 0 ←→ 0 < x 2 ? 3

: (1 ? a1 + b1 )(1 ? a2 + b2 )(1 ? a3 + b3 ) = 4(1 + x3 )2 4(1 + (2 ? √ √ 3)3 )2 = 5616 ? 3240 3

~K 1.15.

ê {Fn }÷vF1 = 1, F2 = 1, Fm+1 = Fm + Fm?1 , m


2,??:

I=
k=0

1 F2k

y?. ?

√ √ 5+1 n 5?1 n 1 ) ?( ) ) Fn = √ (( 2 2 5 √ 5+1 K 2

·?Pa =

√ 1 5 = 2n F2k a ? a?2n √ √ 5x 1 1 1 = 2 = 5( ? 2 ) F2k x ?1 x?1 x ?1

2-a2 = x

n

:

:


I=
k=0

√ 1 =I= 5 F2k

√ √ 1 1 5 7? 5 )= ( 2n ? 2n+1 = a a?1 2 a ? 1 k=0


~K 1.16. a, b, c, d ∈ [0, π ],?…÷v sin a + 7 sin b = 4(sin c + 2 sin d); cos a + 7 cos b = 4(cos c + 2 cos d) y?:2 cos(a ? d) = 7 cos(b ? c) y?. ~K 1.17. y?(4 cos2 9o ? 3)(4 cos2 27o ? 3) = tan 9o y?. |^cos 3x = 4 cos3 x ? 3 cos x ? cos 3x = 4 cos2 x ? 3-x = 9, x = 27?? y cos x

7

~K 1.18. 3n /ABC ?,sin A + sin B + sin C

1,y?:

min{A + B, B + C, C + A} < 30o y?. ?”A B C Ke??Iy?B + C < 30o 1 5? A sin A > 2 sin A ? sin A < 1 2

60K7,A > 150 ? B + C < 30

~K 1.19. 3n /ABC ?,A ? B = 120o , R = 8r,?cos C y?. 5? ? 4r A B C = sin sin sin R 2 2 2 K 1 A B C A?B A+B C = 2 sin sin sin = (cos ? cos ) sin 16 2 2 2 2 2 2 2‘\A ? B = 120 ? 1 C C 1 ( ? sin ) sin = 2 2 2 16 ) sin C 1 = l 2 4 cos C = 1 ? 2 sin2 7 C = 2 8

~K 1.20. y?cos 1o ??nê y?. 1ecos 1o ??nê,@od ú? cos 2o ???nê,e?|^

cos(n + 1) + cos(n ? 1) = 2 cos n cos 1 `??±??4íe ,ù ?cos 60o ??nêg?

~K 1.21. a, b, c ∈ Ry? (ab + bc + ca ? 1)2 y?. -a = tan x, b = tan y, c = tan z K ? ( tan x tan y ? 1)2 (a2 + 1)(b2 + 1)(c2 + 1) d 1 cos2 x cos2 y cos2 z 1

? ((ab + bc + ca ? 1) cos x cos y cos z )2

5? (ab + bc) cos x cos y cos z = sin y sin(x + z ), (ca ? 1) cos x cos y cos z = ? cos y cos(x + z )K ? (cos(x + y + z ))2 1

8

~K 1.22. y?e?? ? | sin a1 | + | sin a2 | + ... + | sin an | + | cos(a1 + a2 + ... + an )| y?. én8B,éun = 15? : | cos a1 | + | sin a1 | 1en ???? y,e?y?n + 1 cos2 a1 + sin2 a1 = 1 1

??: d8Bb ,?Iy? | cos(a1 + a2 + ... + an )|

| sin an+1 | + | cos(a1 + ... + an+1 )| -sk ?ak ?ê ,5?

| cos sn | = | cos(sn+1 ? an+1 )| = | cos sn+1 cos an+1 + sin sn+1 sin an+1 | | cos sn+1 cos an+1 | + | sin sn+1 sin an+1 |le| cos sn+1 | + | sin an+1 | ¤8B 2? ~K 1.23. l?m[ 2 √ √ 6 √ , 2+ 2 √ 6 ] ? o?ê,y?7,?3ü?êa, b÷v 2

|a 4 ? b2 ? b 4 ? a2 | y?. -a = 2 sin x, b = 2 sin y K |a

4 ? b2 ? b 4 ? a2 | = 4| sin(x ? y )| ? | sin(x ? y )|

sin

π 6

5? ^??m??[?2 sin 15, 2 sin 75]Kx, y ∈ [?15, 75] ¤n??m[?15, 15], [15, 45], [45, 75]? o?ê7,kü?3???mS,d??? ~K 1.24. n /ABC S ? u120o ,y? √ cos A + cos B ? cos C 3 >? sin A + sin B ? sin C 3 y?. En /ABC ,PA = 120o ? A, B = 120o ? B, C = 120o ? C Kd^??A B C ?? √ sin A + sin B > sin C ←→ 5? sin A + sin B ? sin C > 0 ~ K1.25. ?ê 0? u30

n /,Kkn ? ?? 3 1 (cos A + cos B ? cos C ) + (sin A + sin B ? sin C ) > 0 2 2 n ¤?? ? 1k? ??? t?? a2004 =

an ÷va1 = t, an+1 = 4an (1 ? an ), n

9

y?. 4í? dan+1 ? 1 = ?(2an ? 1)2 ‰“?2an ? 1 = cos bn K sin2 bn = sin2 du4x(1 ? x) ∈ (0, 1) ?0 x bn+1 ? bn+1 = 2bn 2

1@o??a2004 = 0Kt ∈ (0, 1)K?” a1 = sin2 a kπ kπ @o a2004 = sin2 22003 a ? a = 2003 ? t = sin2 2003 2 2 k = 0, 1, 2..., 22002 ? 22002 ???t? ~K 1.26. a0 = √ 2+ √ 3+ √ 6,?…an+1 = a2 n?5 , n > 0y? 2(an + 2)

an = cot y?. ?bn = an + 2 :

2n?3 π ?2 3 b2 n?1 2bn

bn+1 = 2‰“?bn = cot cn ?

cot cn+1 = cot 2cn 25? :
π π π √ √ √ cos 24 2 cos2 24 1 + cos 12 1 + 42 + 46 π √ √ = = 2 + 2 + 3 + 6 = a0 + 2 = = = cot π π π π 6 24 sin 24 2 sin 24 cos 24 sin 12 ? 2 4 4 √ √

c0 =

2n?3 π π Kcn = ‘\ ¤??‘ 24 3

π ~K 1.27. a, b, c ∈ [0, ],y? 2 sin a sin(a ? b) sin(a ? c) sin b sin(b ? c) sin(b ? a) sin c sin(c ? a) sin(c ? b) + + sin(b + c) sin(c + a) sin(a + b) y?. |^n ?? ú? sin(x + y ) sin(x ? y ) = K ? d sin a(sin2 a ? sin2 b)(sin2 a ? sin2 c)
cyc

0

cos 2y ? cos 2x 1 ? 2 sin2 y ? 1 + 2 sin2 x = = sin2 x ? sin2 y 2 2

0

-sin a = x, sin b = y, sin c = z

dy? x(x2 ? y 2 )(x2 ? z 2 )
cyc

0

?” x

y

z K? x(x2 ? y 2 )(x2 ? z 2 ) 0; z (z 2 ? x2 )(z 2 ? y 2 ) y (z 2 ? y 2 )(y 2 ? x2 )

?\y. 10

ps :???·?k:a, b, c > 0, k ∈ RK? ak (a ? b)(a ? c)
cyc

0

1 2 ps:??? ka, b, c > 0, X, Y, Z ?'ua, b, c K??k =

?ê,??a, b, cüN? 0

X, Y, Z üN?k:

X (a ? b)(a ? c)

π ~K 1.28. x1 , x2 , ..., x10 ∈ [0, ]÷vsin2 x1 + sin2 x2 + ... + sin2 x10 = 1 ,y? 2 3(sin x1 + sin x2 + ... + sin x10 ) y?. ^? d sin2 x2 + sin2 x3 + ... + sin2 x10 = cos2 x1 Kd…?? ? 3 cos x1 ?n ? 10??f?? ? y sin x2 + sin x3 + ... + sin x10 cos x1 + cos x2 + ... + cos x10

π 5?(1998U SAM O)-a0 , a1 , ..., an ∈ (0, )?…?? 2
n

tan(ak ?
k=0

π ) 4

n?1

y?
n

nn+1
k=0

y?. -bk = tan(ak ?

π )K?1 < bk < 1?… 4 1 + bk (1 ? bl )
0 l=k n

Kd??? ?: ? (1 ? bl )
0 l =k n

?1/n (1 ? bl )?

n?

0 l =k n

Kk?
n n n

nn+1
k=0

(1 ? bk )
k=0

←→
k=0

1 + bk 1 ? bk

nn+1

11

n

5???mk ∈ R+ , p

2, p ∈ N ÷v
k=1

1 = 1 y? 1 + mp k (n ? 1) p
n

m1 m2 ...mn ~K 1.29. ?¤kn

?ê|(a, b, c)÷va2 ? 2b2 = 1, 2b2 ? 3c2 = 1, ab + bc + ca = 1

y?. ^?? (a2 + 1) = 2(b2 + 1) = 3(c2 + 1)2‰“?a = cot A, b = cot B, c = cot C sin A : sin B : sin C = 1 : Kn>'???, d?? n /Kc = 0, a = ~K 1.30. b n ( /ABC ?,y? 4 √ √ 2: √ 3

1 2, b = √ 2

cos A 2 cos B 2 cos C 2 ) +( ) +( ) + 8 cos A cos B cos C cos B cos C cos A

y?. 3i\? ? x2 + y 2 + z 2 ? x= cos B cos C cos A ,y = ,z = cos B cos C cos A LHS 2 2xy cos C + 2yz cos A + 2zx cos B :

cos A cos B cos B cos C cos C cos A + + cos C cos A cos B cos B cos C cos A ? 4 cos2 A + cos2 B + cos2 C

e?23i\?

?? x =

2 (?? ?

cos A cos B cos B cos C cos C cos A + + cos C cos A cos B

cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1 ? ? y ~K 1.31. an ??O êê
n

,?yeén ak sin kx|

1, nan A(1 + π )
m

AKé??

ên

1, x ∈ Rk

|
k=1 n

π y?. ?” 0 < x < π -m = [ ]K x @o???:
m

n

ak sin kx =
k=1 m k=1

ak sin kx +
k=m+1 m k=1

ak sin kx

0
k=1

ak sin kx
k=1

A sin kx k

A kx = Aπ k

12

k

,????-sk (x) =
j =1

(n+1)x sin nx 2 sin 2 Kk sin jx = sin x 2

|Sk (x) ? Sm (x)| = | @o?
n

(k+1)x (m+1)x x +1)x sin kx ? sin mx cos (2k+1) ? cos (2m4 2 sin 2 2 sin 2 4 | = | | sin x 2 sin x 2 2

1 | sin x 2|

n?1

|
k=m+1

ak sin kx| = |
n?1

(ak ? ak+1 )(Sk (x) ? Sm (x)) + an (Sn (x) ? Sm (x))|
k=m+1

(ak ? ak+1 )
k=m+1

1 am+1 1 x + an x = sin 2 sin 2 sin x 2

A

ü??\? ?
n

5? ??Uy?·??± é???êx??? ên,y?

?r

(?|
k=1

ak sin kx|

√ 2A π ?

Xe·K:

n

|
k=1

sin kx | k

√ 2 π

~ K1.32. a, b, c, d? ? ? ?xkf (x) > 0y?f (x) < 3 y?. ?I5?

? ~ ê,ef (x) = 1 + a sin x + b cos x + c sin 2x + d cos 2xé ?

f (x ?

2π 2π ) + f (x) + f (x + )=3 3 3

5?a, b, c, d???0, f (x) = a + b cos 2x + c sin 5x + d cos 8x b f (t) = 4aé? ?êt¤á,y??3?ês? f (s) < 0 y?. -g (x) = be2ix ? ice5ix + de8ix Kf (x) = a + g (x)K g (x) + g (x + K f (x) + f (x + 1. a < 0K s = t 2. a = 0duf (x)???0…±? w,?3s? f (s) < 0 3. a > 0K-x = tk 2π 4π ) + f (x + ) = ?a < 0 3 3 K?>?7,– ?k???ê??K f (x + 2π 4π ) + f (x + ) = 3a 3 3 4π 2π ) + g (x + ) = g (x)(1 + e2iπ/3 + e4iπ/3 ) = 0 3 3

13

2.
~K 2.1. ? ??ê ??ê,K?3n?

ê?1
ü? ?a, b ? ab?1?

êd? u2, 5, 13y?8?{2, 5, 13, d}??±é

y?. ?y,1e??

êx, y, z ÷v:

2d ? 1 = x2 , 5d ? 1 = y 2 , 13d ? 1 = z 2 @ow,x??ê, x = 2k ? 1K:2d ? 1 = x2 ? d = 2k (k ? 1) + 1ù`?d???ê,@oy, z 7

?óê,Py = 2m, z = 2nK: 5d ? 1 = 4m2 , 13d ? 1 = 4n2 ? 2d = (n ? m)(n + m) dun ? m, m + n??óê d?óê,g?! ~ K2.2. a, b, c, d? ? ê,0 < a < b < c < d,? …ad = bc,1 ea + d = 2k , b + c = 2m ?

?k, m?? ê,y?a = 1 y?. dd > c > b > a > 0, ad = bc : (a ? d)2 + 4ad > (b ? c)2 + 4bc ? (a + d)2 > (b + c)2 ? a + d > b + c (?a + d = 2k , b + c = 2m :k > m,rd = 2k ? a, c = 2m ? b‘\ad = bc

a(2k ? a) = b(2m ? b) ? (b + a)(b ? a) = 2m (b ? a2k?m ) dub ? a2k?m ??ê,b + a, b ? a??4n + 2.??4n. b ? a = 2f, b + a = 2m?1 e??e, f ?êef = b ? a2m?1 5? ef = b ? 2k?m a e = 1, : b + a = 2m?1 , b ? a = 2(b ? 2k?m a) ?~ : 2a = 2m?1 ? 2b + 2k?m+1 a ? 2m?1 = 2k?m+1 a ? a = 1 :

b ? 2a < b ? a = 2f

~K 2.3. y?. 8B E k = 2?:

n?

ê,k ?? u2

ê,y?nk ?±L?¤n??U?ê ?

n2 = 1 + 3 + 5 + ... + (2n ? 1) b k = m?k: nm = (a + 1) + (a + 3) + ... + (a + 2n ? 1) = na + n2 14

K nm+1 = nm ×n = (na+n2 )n = n3 +n2 a = n2 +n(n2 +na?n) = 1+3+5+...+(2n?1)+n(n2 +na?n) ·?Pb = n2 + na ? nK: nm+1 = (1 + b) + (3 + b) + ... + (2n ? 1 + b)

~ K2.4. y?é?? ??ê ? √ y?. Pa = [ n],Ka2 Pb = [ n ? a2 ]Kb2
2 2

√ ên?U3?m n, n + 3 4 n + 1 ?é

??ê,§UL??ü?

n < a2 + 2a + 1 ? a2 n ? a2 2a

n

a2 + 2 a

m = a + (b ? 1) Km?ü

??ê??,?…m > n,qk: 2b + 1 √ 2 2a + 1 √ √ √ 2 24n+1<34n+1

m ? n = a2 + b2 + 2b + 1 ? n √ ù`?m ∈ n, n + 3 4 n + 1 ~ K 2.5. k ??u1 ??

ê,y??U2k × k

??L?W\êi1, 2, ..., k 2 ±?

?1

? ê ???2

y?. b ?3÷v?? W\?Y, ¤kk ?1??? 2a qduz?1 ê, ?2a ?, ê ???2a

?2a Kk:

1 1 + 2 + .. + k = k (k + 1) 2 ê,¤±2a AT?¤kê

k? ? ê

1 ?1 + 2 + ... + k 2 = k 2 (k 2 + 1) 2 1 2 2 1 a ? ?, k (k + 1)? ? ê,? ¤ á, k ? ó ê?2 A T ? k 2 , ? ù 2 2

1 k (k + 1)g?,?d??3÷v?? W\?? 2 ~ K2.6. r8??? ê?OI3 ?N l??:?,23c?I?ü?? ? êi??? u¤ ?: ?

?ú ê,??U?k?Uc? ¤kêi????:¤kêi??? y?. |^ a = b?3 (a, b) a+b 12?? ?,??

?¤k?:?

kc?êi??,

?¤á …= z|? ?:???,?? ê ? ú ê,y?σ (n) 2n?1

,ùw,??U

~K 2.7. ^σ (n)L?cn? y?. ky? 0 ¤k?êl ,???, k n ? 1?k

n! |σ (n) k !(n ? 1 ? k )! k ??g?p1 < p2 < ..., ?K êak ÷vpa k
aj

ak +1 k n < pk , Kpa k |σ (n)

n! ?pk k !(n ? 1 ? k )! k pj k

gê?
j =1

[

n pj k

]?[

k pj k

]?[

n?1?k pj k

]

qk:

[

n pj k

]?[

]?[

n?1?k pj k

]

n pj k

?

k pj k

?1+

1 pj k

?

n?1?k pj k

?1+

1 pj k

<2

15

?[ = n! ?pk k !(n ? 1 ? k )!

n pj k

]?[

k pj k

]?[

n?1?k pj k

]

1

gê??Lak l k: 1 n
n?1 k=0

n! |σ (n) ,@o: k !(n ? 1 ? k )!

σ (n)

n! = 2n?1 k !(n ? 1 ? k )!

3.

ê?2
ê??

~K 3.1. y?e??§k??|? ê knê),??{a}L?a {x3 } + {y 3 } = {z 3 } y?. 5? : 3 5
3

+

4 5

3

=

6 5

3

?1

3 4 6 ¤±x = , y = , z = ??|),ù ·? E: 5 5 5 3 4 6 x = (125k + 1), y = (125 + k ), z = (125 + k ), k ∈ N + 5 5 5 ÷v?? éu?? {xn } + {y n } = {z n }, n ∈ N + êλn = kmn + 1@o: 1 1 mn?2 1 + nλ + N = + +k+N 2 n 2 n m m mn m2(n?1)

? m > 1? (m, n) = 1K?3 1 + λmn?2 m2 ??N ??? ê,ù { : 1 m2
n n

=

}+{

1 m

n

}={

1 + λmn?2 m2

n

}

~K 3.2. ?¤kg,ê|(x, y, z )? y ??ê,?…y, 3?? ?z ,x3 ? y 3 = z 2 y?. 5? :(x ? y )(x2 + xy + y 2 ) = z 2 , x2 + xy + y 2 ? (x + 2y )(x ? y ) = 3y 2 ù x ? y, x2 + xy + y 2
2 2

ú êò

?y, z

?Uk??ú ê?1

@ox ? y ?x + xy + y ??

??ê,·?Px = a2 + y ‘\ :

4(x2 + xy + y 2 ) = 4(a2 + y )2 + 4(a2 + y )y + 4y 2 = (2a2 + 3y )2 + 3y 2 ?
2

??ê

Pm = 4(x2 + xy + y 2 ), n = 2a2 + 3y K:m2 = n2 + 3y 2 ? (m ? n)(m + n) = 3y 2 (?y ??ê ?kXe??: 16

1. m + n = 3y 2 , m ? n = 1 K2n = 3y 2 ? 1 ? 4a2 = 3y 2 ? 6y ? 1ü>mod3 2. m + n = 3y, m ? n = y Kn = y = 2a2 + 3y ? y = a2 ?y ??êg? 3. m + n = y 2 , m ? n = 3 K2n = y 2 ? 3 ? 4a2 = y 2 ? 6y ? 35? y 10 ?? g?

(y ? 3)2 > y 2 ? 6y ? 3 = (4a)2 > (y ? 4)2 d?w,?), ?Uky 9? ??)y = 7, x = 8, z = 13

~K 3.3. ?¤k ê|(x, y )÷v x3 = y 3 + 2 y 2 + 1 y?. d2y 2 + 1 > 0 x3 > y 3 ,5? : x3 + y 2 + 3y = (y + 1)3 1ey 2 + 3y > 0,K(y + 1)3 > x3 > y 3 d??) 1ey 2 + 3y 0=?3 y 0 y :(x, y ) = (1, 0), (1, ?2), (?2, ?3)

~K 3.4. ?¤k ê|(x, y )÷v y 3 ? 1 = x4 + x2 y?. ?” x 0,5? :y 3 = x4 + 2x2 + 1 ? x2 = (x2 + x + 1)(x2 ? x + 1)

?…(x2 + x + 1, x2 ? x + 1) = (x(x + 1) + 1, 2x) = 1, ? x2 + x + 1 = m3 , x2 ? x + 1 = n3 ?…5? m3 > x2 ,Kk: x2 ? x + 1 = n3 (m ? 1)3 < m3 ? 3m2 + 3m = x2 + x + 1 ? 3m2 + 3m x2 + x + 1 ? 2x4/3 ? x1/3 ?Ukx = 1, 0 ~K 3.5. ?¤k ?)(x, y ) = (0, 1) ê|(x, y, z )??÷v xy + y x = z y xy + 2012 = y z +1 1

17

y?. 1ex??ê,Kxy + 2012??ê,@oy z ??ê,l y ??ê,z ?óê é1???fü>mod4 é1 ??fü>mod4 ê, ?8? ?2012 :x + y ≡ z y (mod4)duz ?óê,y ?u1, :x ≡ y (mod4)g? ??óê, y > 2?xy ?8 ê,d?y z +1 ????8 )g?,l y = 2 x2 + 2x = z 2 ? (z ? x)(z + x) = 2x Pz ? x = 2u , z + x = 2v Kv > u, u + v = xKv > x = 2v?1 ? 2u?1 ?Ukx = 2, 4, 6, 8, 10 ~K 3.6. ?¤k x ,u 2 v?1 : 2x x + y ≡ 0(mod4)

x7,?óê, x = 2x1 ,Ky, z ?

2v?1 ? 2v?2 = 2v?2

2x/2?2 ? 16x2

y?x = 6, z = 10, y = 2 ên? e?? ê,??[x]L???Lx n2 + 1 √ 2 [ n] + 2 ?? ê

y?. Pn = m2 + k ??0

k

√ 2m, m = [ n],@o:

(m2 + k )2 + 1 (k ? 2)2 + 1 2 = m + 2 k ? 2 + m2 + 2 m2 + 2 5? :(k ? 2)2 + 1 (2m ? 2)2 + 1 = 4m2 ? 8m + 5 < 4m2 + 8 ?UkXe??

1. (k ? 2)2 + 1 = 3m2 + 6 d?éü>mod3 g?

2. (k ? 2)2 + 1 = 2m2 + 4 d(k ? 2)2 ? 2m2 = 3ü>mod3 ? ?m>?k?g 3. (k ? 2)2 ? 2m2 = m2 + 2 d(k ? 2 ? m)(k ? 2 + m) = 1 ? k ? 2 ? m = k ? 2 + m = 1 ? m = 0, k = 3ù ?k n??) ~K 3.7. (?¤±n ê|(a, b, c)? a + b + c?a, b, c ? ú ê 2mg? 3,g? :(k ? 2)2 + m2 ≡ 0(mod3)?Ukk ? 2 ≡ m ≡ 0(mod3)

y?. a, b, c? ?¤á,e ? ? ?” a b cKa + b < 2c c < a + b + c < 3c?…c|[a, b, c] ?[a, b, c] = 2c=c = a + b b|2a(?a b?Ukb = a, 2a ?[a, a, 2a] = 2a = a + b + c = 4a db|[a, b, c] = 2a + 2b

1eb = a,K:(a, b, c) = (a, a, 2a) ?????

1eb = 2a,K:(a, b, c) = (a, 2a, 3a)d?[a, 2a, 3a] = 6a = a + b + c???? ê|?{a, b, c} = {k, 2k, 3k }??k ∈ N + 18

~ K3.8.

n ∈ N + , 1, 2, ..., n

?

ú

ê P ?f (n);

n n n , , ..., 1 2 n

?

ú

êP

?g (n)?y:(n + 1)g (n) = f (n + 1) y?. ·?ky? f (n + 1)|(n + 1)g (n) n n?1 n =n ¤±m|(n + 1) k k?1 m?1 l f (n + 1)|(n + 1)g (n) éu1 m nkk e?y? (n + 1)g (n)|f (n + 1) éu?êp,·???r÷vpr qkp3(n + 1) 5? : [ l :
r

m|(n + 1)g (n), 1

m

n+1

n + 1 < pr+1 ,
r

pr |f (n + 1)

n ? m

gê?
k=1

[

m n?m n+1 ]?[ k]?[ k ] k p p p

n+1 m n?m 1 m n?m 1 1 ]?[ k]?[ k ] = k +{ k}+ < 2(1 ? k ) + k < 2 k k p p p p p p p p

[
k=1

n+1 m n?m ]?[ k]?[ k ] pk p p

r

(n + 1)g (n)|f (n + 1),l (n + 1)g (n) = f (n + 1)

~K 3.9. -m, n(m < n)?

ê,1e n2 + m 2 p= √ n2 ? m2

??ê,y?p ≡ 1(mod8) y?. 1ep?4k + 3.?ê,Kdp|n2 + m2 7,kp|n, m n = pv, m = pu‘\ :

v 2 + u2 1= √ v 2 ? u2 d??k)v = 1, u = 0g? n2 ? m2 = k 2 Kn2 + m2 = k 2 + 2m2 p=k+ 1ep = 2 ?4k + 1. 2m2 k

:2m2 + k 2 = 2k ? 2m2 + (k ? 1)2 = 1?) g?{,K?U?8k +1, 8k +3., 8k +3? , p ≡ 1(mod8) 19

1ep?4k +1.?ê,Kp|k 2 +2m2 d?-2?modp

~K 3.10. -m, n?ü?p?

?ê,y?e??óê,??[x]L???Lx
n?1 2

?? ê

[
i=1

mi 1 + ] n 2

1 y?. |^[x + ] = [2x] ? [x] 2
n?1 2

:
n?1 2

k=1

mk 1 + ]= [ n 2

k=1

2mk ]? [ n

n?1 2

[
k=1

mk ] n mk ] n [
k<n/2,2 k

n?1

=
k=2,2|k

2mk [ ]? n [

n?1 2

[
k=1

=
k>n/2,2|k

2mk ]? n 2mk ]? n 2mk ]? n 2mk ]? n [

mk ] n m(n ? k ) ] n mk ] n mk ] n

=
k>n/2,2|k

[

[
k>n/2,2|k

=
k>n/2,2|k

[

[m ?
k>n/2,2|k

=
k>n/2,2|k

[

m?1?[
k>n/2,2|k

=2
k>n/2,2|k

m?1 mk ]? n 2

ù`? ??óê. 5: 1 K?$^ [x + ] = [2x] ? [x]?Hermite? ? 2 [x] + [x + y?. -f (x) = [x] + [x + f (x + A~:

1 2 n?1 ] + [x + ] + ... + [x + ] = [nx] n n n

1 n?1 ] + ... + [x + ] ? [nx]K: n n

1 n?1 1 ) = [x + ] + ... + [x + ] + [x + 1] ? ([nx] + 1) = f (x) n n n x< 1 ?w,k: f (x) = 0 + 0 + ... + 0 ? 0 = 0 n ê,y? pn k pn ≡ k (modp)

1 ?df (x)± ?±?, 30 n é???ê ?¤á ~K 3.11. -p??ê,k ?

20

y?. ?kd¤ê ?n

:ap = (ap )p

k

k?1

= ap

k?1

= ... = a(modp)

Pa = kpn e?? (1 + x)a ?m?: (1 + x)a ≡ (1 + xp )k (modp) ?xp
n n



: kpn pn ≡ k 1 = k (modp)

~K 3.12. é??g,ên

14y?e?,??[x]L???Lx

?? ê

√ √ √ √ √ √ [ n + n + 1 + n + 2 + n + 3 + n + 4] = [ 25n + 49] y?. Pf (x) = √ x

???,duf (x)?à,djensen? ??: √ n+ √ n + 1 + ... + √ n+4<5 : x n+2+ √ ? 2 n+2
x 0

√ n+n+1+n+2+n+3+n+4 =5 n+2 5

,???,d‘??{‘ √

Vú? √

n+2+x=

(x ? t)dx 4 (n + 2 + t)3

: √ 5 n+k 5 n + 2 ? 3/2 < 4n k=0 √ 5? k: √
4 4

√ √ : 25n + 50 ? 25n + 49 = √

1 1 5 √ > √ > n?·? 4n3/2 25n + 50 + 25n + 49 2 25n + 50 √ √

25n + 49 <
k=0

n+k <

25n + 50

? y ~K 3.13. -a, b? y?. é?? ê÷va|b2 , b3 |a4 , a5 |b6 , b7 |a8 , ...y?a = b (4n + 2)υp (b)

ênka4n+1 |b4n+2 , b4n+3 |a4n+4 ,K(4n + 1)υp (a) υp (a) υp (b) 4n + 2 4n + 1

-n → ∞

:υp (a)

υp (b) (4n + 4)υp (b) 4? υp (a) υp (b)

?nd(4n + 3)υp (a)

υp (a) = υp (b),=a = b ~K 3.14. y?é?? êa, b, ce?¤á,??[a, b, c]L?A?ê ??ú ê [a, b, c]2 |[a, b][b, c][c, a] 21

y?. é???êp,Px = υp (a), y = υp (b), z = υp (c)K: υp ([a, b, c]2 ) = 2 max{x, y, z }, υp ([a, b][b, c][c, a]) = max{x, y } + max{z, y } + max{x, z } ?K duy?: max{x, y } + max{z, y } + max{x, z } ?” x y z K?? dy 0¤á ê 2 max{x, y, z }

~K 3.15. y?éu??g,ên,n!?e?
n?1

(2n ? 2k )
k=0

y?.

??fp,·??Iy?
n?1

υp (n!) éup = 2k:

υp
k=0

(2n ? 2k )

υ2 (n!) = n ? s2 (n)

n?1

??Sp ?rn! ¤p?? ??ê ?êè??,,????
n?1 n?1 n k n?1

υ2
k=0

(2 ? 2 )

=
k=0

υ2 2 ? 2

n

k

=
k=0

2k = 2k+1 ? 1

n?1

d??? y,e? p > 2,Kk:p|2p?1 ? 1,@o:
n?1

2n ? 2k = 2
k=0

n(n?1) 2

n

(2k ? 1)
k=1

@o???:
n?1 n?1

υp
k=0

(2n ? 2k )

=
k=0

(2n?k ? 1)
1 k(p?1) n

υp (2k(p?1) ? 1)

[

n ] p?1

,???: υp (n!) = l υ2 (n!) [ n ] p?1
n?1

n ? sp (n) p?1

n?1 n < p?1 p?1

υp (n!)

υp
k=0

(2n ? 2k )

22

~K 3.16. -P (x) = xn + an?1 xn?1 + ... + a1 x + a0 ??‘???a0 = 0 ?… |an?1 | > 1 + |an?2 | + ... + |a1 | + |a0 | y?P (x)?? ~ K3.17. -f (x) = an xn + an?1 xn?1 + ... + a1 x + a0 ? 0, |ai | H, i = 0, 1, ..., n ? 2??H ?,? ê, ????E":α?o? ??? ?o÷v: |α| < ~K 3.18. f (x) = (x ? 1)2 (x ? 2)2 ...(x ? 2013)2 + 2014 y?f (x)3knê??? ~K 3.19. -a1 , a2 , ..., an ?n??? ê,Ke?3knê??? 1+ √ 1 + 4H 2 Kéuf (x) Xê?‘?,??an 1, an?1

(x ? a1 )2 (x ? a2 )2 ...(x ? an )2 + 1 ~K 3.20. f (x) = (x ? 1)4 (x ? 2)4 ...(x ? 2013)4 + 2014 y?f (x)3knê??? ~ K3.21. Pf (x)??Xê?1 – ??g,ê),y?deg f = 1 ~K3.22. -p?????ê,p?1

Xê?‘?,1eéu??g,ên,?§f (x) = 2n ok

f (x) =
i=1

i p

xi?1

1. y?f (x)

x?1

?;f (x) (x ? 1)2

? …= p ≡ 3(mod4) ? ? (x ? 1)3 ?

2. y? p ≡ 5(mod8)?f (x) (x ? 1)2 ~K 3.23. h, k ?ü?p? ê,k?1

S (h, k ) =
r=1

r k

hr hr 1 ?[ ]? k k 2

y?? 12hkS (h, k ) + 12khS (k, h) = h2 + k 2 ? 3kh + 1

23


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