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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:
- [' S( c# X0 H6 h" iLet n >1 be an integer 8 ~# }+ X5 \+ j% N( g3 I( c2 h
Basis: (n=2)
) M9 M2 f% s7 @# U4 r6 J 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3( R2 Z( x" d) [, `' l# [1 X: G
% S1 _7 c/ ]2 _$ c6 i% }Induction Hypothesis: Let K >=2 be integers, support that
7 P8 L2 m! W! w K^3 – K can by divided by 3.' M, Z" \, k. m/ ~5 q3 s% b
0 K& _, s) X6 n' V2 q" ]
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3$ R$ g& o( o9 D; ~7 e0 X
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem' S( A" K* f9 G" w$ R: s* c
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)* s- F2 A, e4 Q7 R
= K^3 + 3K^2 + 2K
5 }' \: b1 C2 m& c5 R = ( K^3 – K) + ( 3K^2 + 3K)' @2 @$ ]# ^, q1 e! j8 W
= ( K^3 – K) + 3 ( K^2 + K), P5 n' X( N, t( A) r1 ^* ~. Z
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
4 v5 F3 F* N( R- P/ WSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)" a, E* c$ Q3 P1 N0 N$ p- t$ b
= 3X + 3 ( K^2 + K)7 ^- D8 I0 U/ m3 [0 f1 n; J) k& c
= 3(X+ K^2 + K) which can be divided by 30 p) @2 ]. j/ g$ V0 w- ^) d
0 S0 z! H. c9 O' S1 `' h
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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9 C! P9 X0 G6 F! H" \[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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