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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)7 f' O+ Z1 `6 t( v5 F% O. J
& f- ?4 T' W5 E3 S: F% pProof:
9 a# |* d' c+ h: w% b- \Let n >1 be an integer t% D5 S: g5 Q4 q& L: D4 l: o8 j) _
Basis: (n=2)
, v7 F/ M/ S' U3 d5 C4 b, v+ _ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 s# [6 h8 K) a1 B# M7 Q
- Q8 X' j' x _ l6 V
Induction Hypothesis: Let K >=2 be integers, support that
) }' n8 P8 j1 {. a4 y K^3 – K can by divided by 3.' a' d+ A% G4 s" n0 e2 [
: Q2 T; I6 I2 x0 K T& jNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
; X( R9 q6 X( @) ]9 ]7 esince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
- Q# C# N- u6 M F6 c/ OThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)$ J7 @3 z7 t* P
= K^3 + 3K^2 + 2K) ]9 J& d& `9 d L7 L* Q; [% B
= ( K^3 – K) + ( 3K^2 + 3K)
5 o5 x& S' O1 P `* | = ( K^3 – K) + 3 ( K^2 + K)- y. ?& V' A# G1 o2 c* f% e
by 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
% j$ j! _' @+ F; F2 V. {" bSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K) K+ t: \" V7 H$ ~, S$ E
= 3X + 3 ( K^2 + K)
& T' i6 h8 j9 x = 3(X+ K^2 + K) which can be divided by 3 M& [1 n' a, m0 S! w
0 o/ y3 \5 R- b }5 f: Q T
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; k4 S6 |9 r5 q3 z
& z0 e5 K- B& L2 d9 r; B
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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