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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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7 x0 U- V. s4 M7 ^! ?Proof:
5 E4 l2 |" w0 Z3 R: WLet n >1 be an integer 5 j. i: |& d; I% q! Z# Z
Basis: (n=2)
6 D! {3 E* M2 k; M 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3+ G' F. b% |6 U1 J; G; l
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Induction Hypothesis: Let K >=2 be integers, support that! e3 Z; W, F) W6 U" T
K^3 – K can by divided by 3.
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# e( z2 V" k/ dNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3' I! G R# l! i
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem- b: X+ K: o) r( d, v8 D/ L6 ?
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
6 x+ {4 w. K7 q2 Q4 p0 [ I = K^3 + 3K^2 + 2K; t, }2 |2 M6 k b- `
= ( K^3 – K) + ( 3K^2 + 3K)
; J( I2 J' ~( ]7 x' _' } = ( K^3 – K) + 3 ( K^2 + K)
" D' c6 L# l9 e0 s! f4 l7 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
9 {9 \6 I4 c4 |0 H- Y6 YSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
8 r) r5 J! G- D; `# w; P+ m = 3X + 3 ( K^2 + K)8 V- t9 v5 J" E/ W$ p; }
= 3(X+ K^2 + K) which can be divided by 3$ c1 N& D! i( \ k" x
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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