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This can by done by Induction
Show that for all integers n >1, n^3 - n can be divided by 3. (Note: n^3 stands for n*n*n)
5 v5 h2 m+ `$ U$ Z; j2 j. ?6 h8 }$ R6 }. {2 F! q; ]
Proof:
2 `- [9 y, B" A) @! C# X6 WLet n >1 be an integer
% s. d/ s* C9 H7 QBasis: (n=2), Z/ V9 C3 w1 S. J1 M
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 d9 t. ?5 p0 D5 o5 X' W# s% p
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Induction Hypothesis: Let K >=2 be integers, support that' C* H6 \4 y6 W2 ~$ l9 W
K^3 – K can by divided by 3.# D2 a0 [2 g1 ^& H* V& Y
; f5 o E* |0 ~+ d$ wNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
8 B# X" `# \$ F6 H4 x4 |since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
+ _" H- b! E0 P2 h* B' EThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)/ T& r3 f2 G# {* D' L' L
= K^3 + 3K^2 + 2K8 K9 @/ z! Z; X/ K( B
= ( K^3 – K) + ( 3K^2 + 3K)
9 K- F* ]4 l9 i9 X! A$ i# F: Y. H = ( K^3 – K) + 3 ( K^2 + K)
9 ?' ~+ j% e; d" ^* I9 y n! hby Induction Hypothesis, we know that k^3 – k can by divided by 3 which means that k^3 – k = 3 X for some integer X>0
5 e& ^- r/ D$ e9 Z9 ?. ySo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
* q7 d; q/ c# p G = 3X + 3 ( K^2 + K)
! A! I" @" \; l4 \& M" Z = 3(X+ K^2 + K) which can be divided by 39 t$ n) n7 z' e2 `. z; E$ K
! p8 i/ Z) ~2 hConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.. u9 A D4 p9 U- V
, S; D' ^3 o! Y+ n$ W0 J3 u[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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