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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)! a/ F) w' n u3 L
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Proof:
( D/ L9 j4 u* Z' z5 j# mLet n >1 be an integer
4 S ^7 i: u# K9 a/ {0 @9 M! oBasis: (n=2)6 {" t6 n9 P5 t$ W, n
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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& l) t( B" G0 t9 A9 m. ZInduction Hypothesis: Let K >=2 be integers, support that
- `/ {# E9 d/ r7 d K^3 – K can by divided by 3.: N1 h1 Z) b' G
+ C [# ^* ]# B
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
/ g2 I( e6 X: s8 x9 Lsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
' D/ E j9 g1 U F" n1 a& CThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)* n3 s, P# ~* t O5 H# O
= K^3 + 3K^2 + 2K
$ ~4 d0 g4 u6 C B$ n5 V, _9 B! c = ( K^3 – K) + ( 3K^2 + 3K)
" N/ [1 u: r. S, f = ( K^3 – K) + 3 ( K^2 + K)8 U& C& Q" z* s' J: Y
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
& S8 n3 V$ f7 w+ g. NSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
; H9 j0 \1 B: @6 h8 [9 t = 3X + 3 ( K^2 + K)
1 u( F1 O8 h% @ = 3(X+ K^2 + K) which can be divided by 3; C4 n, Z% {& W% o
$ o }, T2 e0 L& w4 ] D
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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6 E7 Z5 X' r, C1 @* Q- I8 q[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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