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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:
7 j: S+ H1 r1 q2 `5 r; @Let n >1 be an integer
; w) r: R) i6 @6 b& dBasis: (n=2)& R! Y- @. v3 X+ t0 I/ I" ^
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3% x1 B- H! F4 z3 r
% Q, y; \' C, iInduction Hypothesis: Let K >=2 be integers, support that% s5 ~4 c. `# V1 O2 _. K
K^3 – K can by divided by 3.5 V* s1 U) s0 u' P
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 E3 [+ ~8 h! x& W8 dsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
" Q# B" {/ G3 e& |8 ~Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ Y. i- g0 E. Z2 q0 L = K^3 + 3K^2 + 2K
4 O8 M/ i) T; W) n) W = ( K^3 – K) + ( 3K^2 + 3K)
3 o0 R! S* @6 ], E: V = ( K^3 – K) + 3 ( K^2 + K)
, G6 W: q3 B. x! }# yby 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! k U( t% E. g0 } F% D+ y+ t
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K) E$ B6 ?+ {4 m. v. V
= 3X + 3 ( K^2 + K)
) J( x: R6 d: q6 Y = 3(X+ K^2 + K) which can be divided by 3! `% _, A( P3 d* u: t u- m
3 w9 b; S* Y& n hConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1., J) `& V' z4 y E
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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