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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)" P( m6 S9 w6 V8 H ^
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Proof:
/ W9 O- L. D1 x8 n8 j7 C3 fLet n >1 be an integer
7 x8 @. ^/ H, a* u6 f6 tBasis: (n=2)% [& y; f, C: F* m3 W! C
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 Q0 t7 v9 o, q2 R$ M; |' r
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Induction Hypothesis: Let K >=2 be integers, support that' P' j& z, e1 g0 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 33 u$ A2 k+ Z; n1 B9 P Z( n
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
- Z& ]. H# ~, w. l+ C& s$ ~7 U2 B% iThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)( K+ H( n% v. M8 j+ E
= K^3 + 3K^2 + 2K
, Z+ i. [3 H4 `* {) x) G- d = ( K^3 – K) + ( 3K^2 + 3K). i8 {1 ^* x" Q2 U8 ^1 H+ a# x0 o
= ( K^3 – K) + 3 ( K^2 + K)
" D; g" L0 ]( H0 I( n8 B( U) bby 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
. A i3 v% P& q9 r9 ?2 RSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
! G6 ?" `7 p. ?" n* @! [ = 3X + 3 ( K^2 + K)7 a# d7 B! i( k6 B. Z, ~9 d
= 3(X+ K^2 + K) which can be divided by 3
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% |: [8 M( u! R) K7 fConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.. R! u$ b) i, Y0 m! i
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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