answer81 - 4]^L 8101 R B ABCDB C B AD B AB x R MB NB MN B R...

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Unformatted text preview: 4*]^L 8101 * R B ABCDB C B AD B AB x R MB NB MN B R BD B x PC B AC B B B AD B B PC B LB PC AC ACMB CLM AM ML = CM 2 R PB B N P A M D B C P M D N L R 2 A AM ML = CM MN B B CB B QB CN B B DA B B RB ML QC RM = = MD QB AM x Q B C MD RM = ML AM ABCDB CM B AD B CN B AB 2 CD CR B B MD RM = CM AM ML = CM 2 1. *oE@+ y CD B CN B *)oE AM ML = CM 1 E3d*8 (ng*p+oE)` *oE@+ 1 2 2. 3. 1 94~*]^L 4. *)oE { 5. 1 71 2 R * 11 51 > 8102 1 AB B M ~u OE OF P ~u* M 11 (1 )1 O B AB B B DB u A B DPB X* C ~Mu* PB AB B u A AC B BC B EBF X* DO, DA, DB, DE , DF (1) B DPB = 90 - DOP = COD (2) B DPF = 1 1 DPB = COD = CAD 2 2 A, P, D, E B 1 (2) B DPF = CAD = CBD 1 B, P, D , F B (3) B CED = DPA (1 A, P, D, E B = CFD (1 B, P, D, F B ) 1 C , D, E , F B ) COD = DPA = CFD (4) 1 C , D , O, F B (5)1 (3),(4)1 C , D, E , F , O B E , F B C , D, E , F , O @ 1 CE CF 1 OE OF *oE@ u F 71 7m @+oEC 51 +ng*poE)` u 8N* {*)oE - 1 P 2 8103 1 @ n @+oE)ng*p B nB * uN 8 1 * uF x2 0,1,4,0,7 (mod 9) 1 x 0, 1 , 2 , 3 , 4 (mod 9) y H (mod 9) 7 n 0,1,4,0,7(mod 9)1 1 n 0 (mod 9) 1 n 1 (mod 3) nB 1 , 4 1 121 1 , 221 4 1 nB 3mB 71* mu F 2 11...122...225 1 111 B 33...35 B 1m111 1 1 1 B m1 3 m1 2 (1 13 8u* N 8931 721 49 13) nB 1,2,...,2010 1 1 nB 3mB 7 1 111 11 1 (10 m - 1) 2 1 10 2 m - 2 10 m + 1 B 99...9800...01 m -11 9 m -11 0 n 0 (mod 9)1 n B 1 1 1,2,...,2010 * uF B B n 0 (mod 9) 1 n 1 (mod 3) 2231 6701 893 1 1 1. n B 9 |C nC 4@*)oE { n B n 0 (mod 9) 1 n 1 (mod 3) 2. 1 n 0 (mod 9) 1 n 1 (mod 3)8 3 E d* (mod 9) 1 n 1 (mod 3) E@+o o* |C 1 3. C * 5 |C 1 71 11 C 1 61 41 . 7 @ R E o C { 4 m n| 91 n0 8104 1 < an > B a1 = 1, a 2 k = a 2 k -1 + 1 , a 2 k +1 = a 2 k + 2 (k B C )1 a1 + 2a 2 + a3 + 2a 4 + + a 2009 + 2a 2010 a12 = 1 2 2 a 2 k = a 2 k -1 + 2a 2 k -1 + 1 a 2 = a 2 + 4a + 4 2k 2k 2 k +1 1 k=1,2,...,1005 C 2 a 2011 = 1 + 2(a1 + 2a 2 + a3 + 2a 4 + + a 2009 + 2a 2010 ) + 5025 1 2 a1 + 2a 2 + a3 + 2a 4 + + a 2009 + 2a 2010 = (a 2011 - 5026) 2 a 2011 B C 1 2 1 a1 + 2a 2 + a3 + 2a 4 + + a 2009 + 2a 2010 = (a 2011 - 5026) (0 2 - 5026) -2513 2 2 a1 = 1, a 2 = -2, a3 = 0, a 4 k = a 4 k +1 = -1, a 4 k + 2 = 0, a 4 k +3 = -2(k = 1,2,...,501) | (% a 2008 = -1, a 2009 = 1, a 2010 = -2, a 2011 = 0 B B a1 + 2a2 + a3 + 2a4 + + a2009 + 2a2010 = -2513 $ | a2k a1 = 1 B a 2 , a3 , , a 2010 8 u H = 3k - 1, a 2 k +1 = 3k + 1(k = 1,2, ,1005), a 2011 = 3016 1 2 1 a1 + 2a 2 + a3 + 2a 4 + + a 2009 + 2a 2010 = (a 2011 - 5026) = (3016 2 - 5026) = 4545615 2 2 u 8H E o p+)oEP@\^* Xng*p+oE)` * y )oE 8105 *oE@+ 0@ @+oESng*p @ B * )oE y ~ n B * m 8H u 2(n + 1) B 2nB m B * u 28 H * u 18 H n + 1B 2nB mB 2(nB 1)1 m 3 2(n + 1) + 1 B 0u 2nB mB 2(nB 1) 11 m 4 S - 1 (S 8H * )oE8 u u S -1 S ( S - 1) B ~ (y n 1 B m 4 y )~ 2~{ 5 8 H u S ( S - 1) hG * u (8 uH 6 hG u ) 6 * hG u * 6 y)oE A A B C D E F 1 * 1. y )oE 1 1 1 1 1 0 B 1 0 2 2 0 C 1 2 2 0 0 * y )oE D 1 0 0 2 2 E 1 0 2 0 2 F 2 2 2 0 0 1 6 5 5 5 5 4 * y 6 )oE 2. | h*)oE 11 3. 4. 5. a 7m @+oE. ` ) E o + p * g n ( 8 1 71 Em 1 61 08 1 41 | h* 4 }* 21 u 0 * 11 11 61 1 61 1 68 1 21 31 41 58 E3d* 6 ~*]^8 94 ...
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This note was uploaded on 03/25/2011 for the course MATH 1232 taught by Professor Dr.jiangzhengchien during the Spring '11 term at Nanjing University.

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