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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)# x/ f4 }$ M% m' [; ]6 r; n
t$ d) q: u& i8 B5 tProof:
$ j8 C* _* s$ Z% o* c7 Z% g! cLet n >1 be an integer : [: z! P: j: ]& J
Basis: (n=2); k `. S1 r, k% r! @0 C
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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3 q; L4 B* E8 w6 t8 I% u- UInduction Hypothesis: Let K >=2 be integers, support that' E' w2 i$ `8 U/ U' D6 V. F
K^3 – K can by divided by 3.$ N ?5 |, [& Y g
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3. J5 t- a) c% x, [3 Y
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem5 L8 D# Q( _+ [, K, R7 M2 o
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1); y$ x+ Y1 n4 I4 O' }
= K^3 + 3K^2 + 2K/ ?; @ f4 P' J3 v- ^$ S: C
= ( K^3 – K) + ( 3K^2 + 3K)
9 A) M8 H1 H- U8 K = ( K^3 – K) + 3 ( K^2 + K)
2 C5 t4 S' d' c. c: E* j- Y) H4 tby 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>00 }3 N: N5 F# u: v
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
- x1 C5 [+ V' |( l- `1 I5 B = 3X + 3 ( K^2 + K). I6 t5 A0 f+ D& |5 t
= 3(X+ K^2 + K) which can be divided by 3' R% A% Z4 H1 f3 W7 w6 @8 Y/ h' a
3 x9 W$ I. ~8 ? }' l
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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