15 - ›r ›r 901501 + › ˜ Ä € + › ˜ Ä € + ›...

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Unformatted text preview: ›r ›r 901501 + › ˜ Ä € + › ˜ Ä € + › ˜ Ä € + › ˜ Ä € ¸ C K È €Ä˜›r ˜›r€ Ä c ˜¸ª B 9 kÄ €›r ˜ 4(1+++….+) r› ˜ Ä € 4(1+++….+)+1>8.9 ⇒ > 1.975 ⇒1− > 0.9875 ⇒0.0125 > ⇒2k > 80 9 kª ¸ ˜c B 7 † 9 1. ª 2Bc.¸ ¸ 9 9 ¸ C ”• B ” † ª 34 ¸ ¸ Bc 9 19 8.99 ª ¸ 1˜ c B 79 ,9 r› =1”Bc˜•p ˜ Ä ¸ !•ª > ?¸ 7 1” 9 9 210 9 ,9 217 9 9 ,9 9 215 9 !ª ¸ B ,¸ c 202 9 901502 a=1+2+…..+109 b=12+22+…..+1029 c=13+23+…..+103 9 d=14+24+…..+104 9 1~10 9 ! ª ¸ 2 Ø Bc S ¸ T €ˆ 1~10 9 ! ª ¸ B cØ T T N• 9 S=1 × 2+1 × 3+1 × 4+…..+8 × 10+9 × 109 T=1 × 2 × 3 × 4+1 × 2 × 3 × 5+1 × 2 × 3 × 6+…..+6 × 8 × 9 × 10+7 × 8 × 9 × 109 9 a,b,c,dBcØ ª¸ S9 T 9 ª ¸ cØ B ª ¸ cØ B S=(a2− 9 T=[a4− 2b+8ac+3b2− b) 6a 6d] 1.9 1,2,…..,9,10 9 xy □ ×Ξ □ Ξ (1+2+…..+10)(1+2+…..+10)BcX !ª ¸ x=y 12+22+…..+10c X B2 9 x≤ y 9 x≥ y9 S={(1+2+…..+10)(1+2+…..+10)− 12+22+…..+102)= (a2− ( b) 2.9 1,2,…..,9,10 9 4 9 x,y,z,9 u □ × □× □× ¬ □ ¬ ¬ ¬ (1+2+…..+10)(1+2+…..+10)(1+2+…..+10)(1+2+…..+10)BcØ !ª ¸ (1)x=y=z=u 14+24+34…..+104 9 (2)x=y=z≠ u (13+23+…..+103)(1+2+…..+10) − 4+24+…..+104)9 (1 2 2 2 2 2 (3)(x=y)≠ (z=u)9 (1 +2 +…..+10 )(1 +2 +…..+102)− (14+24+…..+104)9 (4)x=y x,z!u ¸ Øc ,ª B 2 2 [(1 +2 +…+102) (1+2+…+10)2− 2+22+…+102)2− 3+23+…+103)(1+2+…+10)+ (1 2(1 4 4 4 2(1 +2 +…+10 )9 4 (9 x × x × y × y x × x × x × u x × x × z × x x × x × x × x) 9 T={(1+2+…+10)4− 4+24+…+104)− 3+23+…+103)(1+2+…+10)− 4+24+…+104)] (1 4[(1 (1 − 2+22+…..+102)2− 4+24+…..+104)]− 2+22+…..+102)(1+2+…..+10)2− 2+22+….. 3[(1 (1 6[(1 (1 22 3 3 3 4 4 4 +10 ) − +2 +…..+10 )(1+2+…..+10)+2(1 +2 +…..+10 )]} 2(1 4 = {a − − d 4[ac− − 2− − d] 3[b d] 6[ba2− 2− b 2ca+2d]} 4 2 2 = {a − − d 4ac+4d− +3d− 3b 6ba +6b2+12ac− 12d} 4 2 2 = {a − b+8ac+3b − 6a 6d} 9 n=109 T=(n− n− n− n(n+1)(15n3+15n2− n− 3)( 2)( 1) 10 8)=1577739 ª ¸ cØ B 1.9 a2=(1+2+3+…..+10)2=(12+22+32+….+102)+2 × (1 × 2+1 × 3+1 × 4+…..+8 × 10+9 × 10) =b+2s 9 s= 2 2.s =(1 × 2+1 × 3+1 × 4+…..+8 × 10+9 × 10)2 =(12 × 22+12 × 32+12 × 42+…+92 × 102)+2(12 × 2 × 3+12 × 2 × 4+…+102 × 7 × 9+102 × 8 × 9) +6(1 × 2 × 3 × 4+1 × 2 × 3 × 5+…..+6 × 8 × 9 × 10+7 × 8 × 9 × 10) (1)b2=(12+22+32+…..+102)2=(14+24+34+…..+104)+2(12 × 22+12 × 32+…..+92 × 102) =d+2(12 × 22+12 × 32+…..+92 × 102) 9 12 × 22+12 × 32+…..+92 × 102= (2)b × s=(12+22+32+….+102) (1 × 2+1 × 3+1 × 4+…..+8 × 10+9 × 10) =(12 × 2 × 3+12 × 2 × 4+….+8 × 9 × 102)+(13 × 2+13 × 3+…..+8 × 103+103 × 9) =(12 × 2 × 3+12 × 2 × 4+….+8 × 9 × 102)+ac− 4+24+…..+104) (1 2× × 2× × 2 9 1 2 3+1 2 4+….+8 × 9 × 10 =bs− ac+d 2 9 (1)(2)9 s =+2(bs− ac+d)+6T 2 ×− ⇒() =+2(b ac+d)+6T ⇒T= 9 1.9 “ T À r0›E@r ! ”~ ~ [2.ظ ºª 17r›E@r ! ~~ @d 7 @ 0= “@ ~ + ”z Ÿ E ~ + @ @d 7 @ 0= ¸ j¸¨ [3.X¸ » !ª !ª .06 X 4¸ [» 58%9 901503 4× 4 + ≅ Ε ∼ ∗ 1 4× 4 ⊗ + 9 4× 4 17 10 16 2 11 6 15 4 14 3 13 7 16 1 10 34 13 14 58 9 12 8 12 59 11 15 26 62 15 11 95 12 8 12 9 85 14 13 43 10 16 13 14 15 11 16 7 1 7 3 4 6 2 10 1 2 4 5 3 6 ¨±[¸ª! !ª (1)h ¸[ ² !ª (2)¨ ¸[ ± 9 n × n ! r› ~E@ 15 × 15¸ π !ª > á5ò‘›r ? 8 9 r› ˜ Ä 7 ˜@²[¸ª! •p ˆ±[¸ª (1) 1 2 3 n2 n2−1 n2−1 n2−2 3 n2 2 1 n2−2 n3 n3 9 n × n p¸ª! >á5ò‘›r ? r› ˜ Ä S˜•• B¸ª! p> 9 <an> <bn> <an>d ªø 1,2,3,….,n2− 2− 2 2,n 1,n 2 <bn>d ªø 1,2,3,….,n − 2− 2 2,n 1,n 22 2 9 <bn>d ªø n ,n − − 1,n 2,….3,2,1 r S˜ • Ä • ˜B¸ªp?! › > 9 =1+n2=2+(n2− 1)=3+(n2− 2)=…..=(n2− 2)+3=(n2− 1)+2=n2+1 9 n× n §ˆ 49 =2(n2+1) (2)9 15 ׸ 15 Ø !ª [ » !ª ¸ [»Ø ª ¸ [4 Ø » r› ˜ Ä € 9 1r › À × 15 ˜[¸ª! »5 ˜ Ä • p 9 <aij>9 (1)9 =2(152+1)=452 =152+1=226 97 89 1 2 4 3 5 6 10 15 a16 a17 9 14 8 13 7 12 11 a61 a71 a81 a82 a87 a11=1,a21=2,a31=4,a41=7,a51=11 … 9 a81=1+(1+2+3+….+7)=1+ =29 9 a12− 11=2,a22− 21=3,a32− 31=4….. 9 a a a a82− 81=9 a 9 a81=29,a82=38,a83=48….. 9 a87=29+(9+10+11+12+13+14)=98 9 8 9 r› ˜7Ä € 98 226− 98=128 >p ª ¸ V r= ˜ Ä • ˜ ?! >› =452− 128=324 Ans9 324 9 1. E r ~ › ! @ r › 2.Ä € ˜ 9 9 !ª ¸ V ; H 3. – 9 ! ª ¸ 11 H V; 9 9 r› ˜ Ä € 9 ª ¸ V; H !5.01 H ª ¸ V; 9 71.57%9 9 9 901504 (1)9 4 x»] (2)9 n x»] (3)9 k,r x ] » 9 k+1 9 4!9 n!x¸ [º k>r>19 r 1À n!=1 × 2 × 3 × …… × n r !9 ¸ [ º xª (1)xª ¸[ º* {4k+1,4k+2,4k+3,4k+4}9 k=0,1,2,……… 9 !ª ¸ [ 4 x º 49 ¸ [4 x º (4k+4,4k+1,4k+2,4k+3)(4k+1,4k+2,4k+3,4k+4)(4k+2,4k+3,4k+4,4k+1) (4k+3,4k+4,4k+1,4k+2) 9 4!9 ª (2)x ¸[ º (3)9 (k+1)(k+2)….(k+r9 1)9 (r9 ¸ 1)x !ª [ º 9 (2)9 (k+1)(k+2)….(k+r9 1)=(r9 1)!m ,9 m 9 k(k+1)(k+2)….(k+r9 1)9 !ª ¸ [ rº x 9 k(k+1)(k+2)…..(k+r9 1)=r! × n,9 n ⇒k × (r9 1)! × m = r! × n ⇒k × m=r × n 9 (k,r)=1 ⇒m 9 r 1 k+1 9 r 1ª°! È[ ¸ r!9 9 1. ^ À nb – 19 29 3….9 n 9 b¸ – ª ¸ [2 È ° r4˜ Ä € › À n!9 !᪠n? pò‘›r 5 ¸> À !ª ¸ 2.È [° !ª ¸ 32 È [° !ª ¸ [ 2 È ° !ª ¸ [ 2 È ° 39 ª ¸[°È ª ¸ [1 È ° 14.28%. ! ª ¸ 3.È [° 9 9 9 ª ¸[° È 9 9 9 9 r› ˜ Ä € ,9 9 901505 @ 7 E ~ @r !ª ¸ > ? pá5ò + › ‘ 0j (0j )9 ª ¸ V 5H @ 7 E ~ @r 9 1.0 2 E ~ + @ 7= 2.0 ! ª ¸ 33 H V5 9 99 99 9 9 ª ¸ V5H 9 9 ª ¸ V5H 9 99 9 99 9 9 99 999 9 9 9 3.9 r ›3˜3Ä € 9 9 9 9 9 9 ª ¸ V5H 9 ª ¸ V5H 9 99 99 9 9 9 74%9 9 9 99 9 !5.18 H ª ¸ V5 ...
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This note was uploaded on 08/14/2010 for the course 數學 乙一 taught by Professor 鄭質明 during the Spring '07 term at Fudan University.

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