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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)
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Proof:
3 r5 v' P7 f Z. [# [# RLet n >1 be an integer
M6 Q Q/ `& f. d; ~% o: c' o! NBasis: (n=2)
7 g5 R, m ~* T1 t 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
8 R: h( U& S: F) B K^3 – K can by divided by 3.4 r1 b |3 J ]) d4 r9 U* y; r! p
- x: Z% e* h& _7 SNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
H# e- Z, W! B# y0 e, A: ~, isince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
( M$ f2 ]* t+ n4 UThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
1 ^6 B/ U0 ~, ~* ?& K3 { = K^3 + 3K^2 + 2K3 C/ \. f& v p5 w" t! h
= ( K^3 – K) + ( 3K^2 + 3K)( {/ ?2 l, r7 F
= ( K^3 – K) + 3 ( K^2 + K)
$ k, f- Z- d! ~8 T Jby 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>09 U% G" E8 `, u9 x' K
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
% }6 C0 I* R6 j G/ Y$ J- N = 3X + 3 ( K^2 + K)6 l# s* I5 O3 H, j$ _' _0 e
= 3(X+ K^2 + K) which can be divided by 3: p' |2 J& q/ N4 _ X- r: {8 {
3 p; ~1 q& S6 H6 c$ t. l) }2 q
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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9 X7 N: Z3 Z! F) J4 E' e# }0 \1 k[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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