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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); r7 V& _# k' _* h. w' H2 \
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Proof:
" A5 L" r( s* ]7 j% {Let n >1 be an integer
3 X) O" L+ y. b. J3 q3 SBasis: (n=2)
# E2 j' R: m' i9 F 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3) I6 A) Q( |8 _6 Q
5 K! N& v8 b) l& P& V4 r( } k6 g7 \
Induction Hypothesis: Let K >=2 be integers, support that
. k. y. C+ }; |, q: O \ K^3 – K can by divided by 3.
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% D$ P1 M0 M* ]- X4 }4 A- ?Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
$ C9 v6 W1 ]4 D0 u' ]since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
7 Q: v/ T# Q/ v! ]! [# F9 `, u: o1 lThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)+ X" q" X- C- y& T2 G- h D2 P
= K^3 + 3K^2 + 2K
& x: V& z/ N% ~8 t' \) n = ( K^3 – K) + ( 3K^2 + 3K)! H. s% O9 [ e: _3 J; [
= ( K^3 – K) + 3 ( K^2 + K)
7 }0 G; l& V+ a9 W1 yby 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& C |5 m b7 n5 i9 x& I
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
' ?8 F7 Y' E/ P, c$ c; ` = 3X + 3 ( K^2 + K)
& E2 Q5 U. b% ` Y6 M0 Z2 g = 3(X+ K^2 + K) which can be divided by 3: x" p; a8 t+ n6 A% h
* A+ ]. U) t' \0 ^9 a1 x3 n" tConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.) c$ T( p! h g- L6 E* j. N
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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