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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)! H' m. t1 G/ b; A* [" c7 Z+ z
% V3 V" M: ]$ Z6 t/ dProof: # E* U; j. m: ], G3 K8 `
Let n >1 be an integer / r5 G3 g+ W" \' d! r' M9 f
Basis: (n=2)
2 w6 q8 q+ g% u! ?# u% b 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that+ J1 ^- A" a- }# B6 y" _* l
K^3 – K can by divided by 3.
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( D* K* C4 G" M/ ~* ~) C6 U% ~Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3; P+ J- W1 ~7 C3 r, C; F* s- q
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
" D9 K- z0 ?" j. B/ a& PThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
9 y# P- M* X" H- O+ B* i( n, Q. b = K^3 + 3K^2 + 2K
w. F, g7 J8 q$ @( D5 `; f = ( K^3 – K) + ( 3K^2 + 3K)
& I( _. k; j/ k2 v = ( K^3 – K) + 3 ( K^2 + K)& G8 V1 e9 c2 x+ c+ G, l7 {8 h7 q
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
/ q% `4 x9 f% I; TSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
& w! `7 a8 `8 f' W& p1 G = 3X + 3 ( K^2 + K)
) `" U0 K9 B* X/ y; f' h8 }2 }8 \ = 3(X+ K^2 + K) which can be divided by 3- g- K5 p# R7 ?0 K3 L+ k
1 d, A' }% Z$ _: sConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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, i0 M+ X' b' S4 K; h4 C[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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