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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)" e" Y) Q0 [6 M; g. f8 V' D7 v
5 m) j; P3 u, Y; I$ X' x% EProof:
" T* Q- Z! W* WLet n >1 be an integer
& j8 x- W3 z1 u$ UBasis: (n=2)
' Z# u1 u3 i; ^2 ~) n5 U1 w 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 that1 ~& M* h/ P1 v% o z
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 3
6 ^4 p8 i9 k" g$ C5 Hsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem% j+ O# ^ J+ L
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
) G4 M! n: b$ Q7 H) l! I, r& G = K^3 + 3K^2 + 2K! `2 o( q! R; p( Q% P# R: q
= ( K^3 – K) + ( 3K^2 + 3K). P" H. y2 z0 a9 z$ `
= ( K^3 – K) + 3 ( K^2 + K)8 V5 _# F3 T& z, O& q% 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
- o- Q/ ?! @* A. f# U5 \So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K) \: v2 N3 @+ Q7 u' \0 @3 }
= 3X + 3 ( K^2 + K)
3 ?3 h1 F6 b4 V* X9 x+ R1 }; I( V, J+ n = 3(X+ K^2 + K) which can be divided by 3) S, d) N+ J' ]7 D; y
# q W( |3 z& n1 uConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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