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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)* ?; i1 m5 W4 D( Y* T
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Proof: ) {' n7 j( k6 l! t
Let n >1 be an integer s& S# T0 z) ^5 N& \0 l
Basis: (n=2)
+ y% }% s ?' w* B% d* n3 x 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
0 t% ~9 Y- E& ~, |4 L) {
7 Q9 {) n. \, f, a2 H3 V0 rInduction Hypothesis: Let K >=2 be integers, support that
% k9 y* g' S) _, S r$ G9 p! D K^3 – K can by divided by 3.2 t7 ?1 p/ l& P( l
: f7 F( O: G3 q; j. }, k
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
% _: z7 y9 j2 G# l9 o3 F/ qsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem2 p# W+ J @( ]! @
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1). P2 R- ^; h) t6 s7 O
= K^3 + 3K^2 + 2K2 F& E: ~. _" m
= ( K^3 – K) + ( 3K^2 + 3K)
+ J/ ] j+ e) Y- ~ = ( K^3 – K) + 3 ( K^2 + K)0 a( z! K4 ?* f# v
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
" J4 I$ m5 X; Q+ m) m# B5 p0 QSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)* W. Y( W0 i& q) D
= 3X + 3 ( K^2 + K)
8 U: k- F" q( g% |* b = 3(X+ K^2 + K) which can be divided by 3
" k' m8 l: s7 Z' X( T( {: W& F' D, M: e3 d6 r6 h1 t0 M' X% G
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1." ?( j- M' ^; I# Q n: Z
, v/ ` W) O9 w[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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