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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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5 y7 g6 R3 |+ O C: jProof:
- G- N' A' @$ z& ZLet n >1 be an integer
5 N. r3 l) r. o- wBasis: (n=2)
8 `8 X+ t% ~5 U8 Z9 r* G, ] 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- h( H0 F1 j& H6 ]) A) ]9 i
K^3 – K can by divided by 3.6 q( ^/ c4 c7 k$ T* G% I
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 38 k# | I# h( L7 |7 v% F8 F
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem2 p9 P+ T; A( I) D/ P& u
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)) N: b4 B/ ^% x
= K^3 + 3K^2 + 2K
. _7 I6 m0 K; |3 O = ( K^3 – K) + ( 3K^2 + 3K)
& b8 S' O' y" t' X. Q/ K/ K! g0 t% O% T: w = ( K^3 – K) + 3 ( K^2 + K); N# _" k2 X3 `6 z3 O( o
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
0 J8 \" u" {. ]1 hSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 b! L1 g4 ~) h% v9 I3 m = 3X + 3 ( K^2 + K)
) {5 L5 ~8 G. R = 3(X+ K^2 + K) which can be divided by 3
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8 k7 w Y3 L' Y9 RConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.5 ~' E" @, z2 _
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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