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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)
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, D; c) C% ^) e( L, K/ NProof: ; z9 J( N' R- n6 W% B
Let n >1 be an integer
$ B4 t. ~1 Q7 G- J' c7 HBasis: (n=2)
% ~ @$ M" @& N3 t/ ^, ]- c* k 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3/ _/ |4 H' G6 q% u/ {4 r' d
& b. V: N/ [5 ]2 a1 V
Induction Hypothesis: Let K >=2 be integers, support that
2 D+ L3 F) {6 d K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 31 _- l3 U# X& h
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem5 m! n h' m, K5 r0 d4 U
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)0 ^9 h1 u1 ~. s! ~
= K^3 + 3K^2 + 2K
" Q* ]. @$ T( D: ]( x5 T, e! k = ( K^3 – K) + ( 3K^2 + 3K)
2 u$ q z8 Q4 d* L, M8 b$ H = ( K^3 – K) + 3 ( K^2 + K)+ g) l+ g8 e$ ^, D% w2 y+ J
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, j G" v% w. Q+ K. U; `+ ~2 d
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
$ o* ] B& [8 f- ]& { = 3X + 3 ( K^2 + K)
0 v& m- C6 u4 _& b = 3(X+ K^2 + K) which can be divided by 3
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& r$ L4 J1 `8 O p* QConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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9 R- R8 m* u8 b3 {9 ~$ M2 a+ I[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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