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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). k* \8 v: }* \: T% B I
8 n: ?2 a; [ _. i$ @' t' FProof:
: K3 p, f: W2 C' {2 h" Y3 H! rLet n >1 be an integer
' }9 x: Q' f- ]4 T0 p! FBasis: (n=2)
& A8 R! A" }. v |* r/ o m; ]5 p 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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$ {5 T0 P6 n- p) k6 K9 I( UInduction Hypothesis: Let K >=2 be integers, support that
4 f8 K/ I" D, _2 N K^3 – K can by divided by 3.1 b) Z8 r% F! p' ~8 B' m. @
A& l* \) F) r4 y6 rNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
- `3 e/ u) B; M4 |# N8 Psince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
0 o& Y' n7 B! Y) `Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1): l* Z- u( r' i1 G% r! _, t2 o
= K^3 + 3K^2 + 2K8 @+ s' v1 u+ b0 F6 a) k
= ( K^3 – K) + ( 3K^2 + 3K)9 G9 U E# D* _/ z2 v$ R
= ( K^3 – K) + 3 ( K^2 + K)
% a/ o. ^, c6 [) ]5 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
- A, s/ c K' hSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
9 z1 i: y* p! a# k# u+ F8 G = 3X + 3 ( K^2 + K)
7 ? [9 Q5 b' J! t8 l& \0 }1 j = 3(X+ K^2 + K) which can be divided by 3
3 |) e9 A1 V4 G4 V1 Y9 h. ]
: Q+ v+ ~8 W v" N2 R( fConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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9 m& b/ K1 T+ R& }5 _! O8 r+ ]3 D[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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