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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)$ y% ]# k7 K0 d- I
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Proof: % o0 U8 p& K& ]) U
Let n >1 be an integer
% i% R6 Y, Y/ s, R; t. H1 IBasis: (n=2)8 F3 ^) n( L" d R6 J
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3/ E9 R3 Q) T: W
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Induction Hypothesis: Let K >=2 be integers, support that
- s% o; h: Y- X2 O K^3 – K can by divided by 3.
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" H. f9 y. t1 g5 g/ }1 A# \7 LNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 36 w. V' S% N2 L5 S2 i* j
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem& o% A* \- i7 w, q" b' B) A- T
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)3 `% f" h0 V- y3 J
= K^3 + 3K^2 + 2K
7 \/ Q8 e. f! T% ?# a/ ]: X$ N = ( K^3 – K) + ( 3K^2 + 3K)/ c* c/ K' `3 |: \$ M! V1 J* B
= ( K^3 – K) + 3 ( K^2 + K)
4 n! d T6 R; [0 S# f Y- Q+ q. 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
1 A3 K' Q( u, Z: v( K# \7 N2 p6 ?So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)9 Y6 I6 Z+ s. N8 ]2 l# j: A- l$ W
= 3X + 3 ( K^2 + K)
- [, s0 e, k9 ~$ k- X1 ]2 z) S0 D = 3(X+ K^2 + K) which can be divided by 3
6 i4 l+ w2 q; [. i+ I) m, r/ y8 L' \8 i- t3 l1 i+ ^+ x4 q
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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