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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:
+ W9 a% C& b) M, s+ K$ W7 ~ zLet n >1 be an integer
+ Q% Y( J2 Y$ B- y \7 E: ]8 ^Basis: (n=2)/ B( Q: z6 _3 X7 T$ N/ G0 H
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3$ [% o4 _ _; w% J2 L6 Q* F
% _- S Z6 ]# z, jInduction Hypothesis: Let K >=2 be integers, support that
- j R& H: @( ], s* i% r# v9 r K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3# T; J- ^# t7 Y! @. k) c# N3 u% l" z
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
2 }6 o! P% B# W( F2 {Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
; a7 m# Z- q! P1 K1 U; J( Q = K^3 + 3K^2 + 2K
0 k2 [: N7 \6 z8 V; j = ( K^3 – K) + ( 3K^2 + 3K)( Y% Q: Q5 E- a9 E
= ( K^3 – K) + 3 ( K^2 + K)
& T4 A/ t6 c! g( a1 i9 D0 Wby 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
$ o* ]; n, Q; `; Y' |$ GSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
* h" @9 X1 b7 F h$ b- z; f = 3X + 3 ( K^2 + K)
8 t( D, _1 D2 n8 L, E. S = 3(X+ K^2 + K) which can be divided by 39 L- X; ~ J- {5 c$ r7 I
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.: R7 `& w- K/ F
! X& s# t* ]; ` v* `[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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