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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)2 G0 ~9 s, I, K+ F8 y/ d: Z
" w& m9 @7 g2 o1 ^) z. LProof:
0 x8 a5 l: U J( e* l$ T& H3 g' v4 |5 eLet n >1 be an integer : f9 r/ Q W; p0 f8 b( f1 ^( w
Basis: (n=2)
k- N# Z/ i5 c+ l, A7 D 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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* \2 v2 D$ r6 X' Q' bInduction Hypothesis: Let K >=2 be integers, support that. e$ A! E* z Z7 X, c; _- v
K^3 – K can by divided by 3.: m% A* s. _3 P# `8 |
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
- R5 v. G D% Esince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem: ]( A4 b* u- n% C+ A! v
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)3 X3 Y$ D! w4 ~' N
= K^3 + 3K^2 + 2K
" l, }5 z( e+ `& B. E# `2 F = ( K^3 – K) + ( 3K^2 + 3K)
7 Y$ l: W- a8 [9 a! p; @* p = ( K^3 – K) + 3 ( K^2 + K)
% x9 m {3 ~' {3 L4 M: cby 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* d4 A' W2 D* e" G3 e( _7 y: w
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) `% }3 l! i. S1 ?6 O = 3X + 3 ( K^2 + K)( k! E; Q: j7 I1 M; b
= 3(X+ K^2 + K) which can be divided by 3
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! `( z% c* l 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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7 j0 U: _ J4 d) F[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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