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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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1 O, ^8 G7 c* O7 p% _Proof:
* S/ z1 L. D% c8 s! mLet n >1 be an integer ) H2 x9 }" E3 v7 Y
Basis: (n=2)5 v$ X# u' j& o4 X$ Y+ u/ m2 n
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3* s1 T+ e. H$ E3 n
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Induction Hypothesis: Let K >=2 be integers, support that
6 g9 m4 \8 H4 t1 f* q W5 u 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
1 S+ k$ v! B6 |) B e- Y# ?since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
! ?! A1 U/ E- r3 t L2 vThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
- M+ ^% H0 |, Z9 L ^7 t = K^3 + 3K^2 + 2K, q' p" w! P G8 v, X
= ( K^3 – K) + ( 3K^2 + 3K)% ?- b7 q E0 T& S6 X
= ( K^3 – K) + 3 ( K^2 + K)
6 c, F2 X# v0 T( o/ y0 W# v9 _- xby 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# K( g9 d- |1 C! y$ {. g
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ L( g- r L( k3 R. K" ?6 M
= 3X + 3 ( K^2 + K) D$ a0 K8 a2 d$ I- [
= 3(X+ K^2 + K) which can be divided by 3. Z' d8 d4 `# f) _& L$ ?! U
3 A! @9 \7 {. l' V9 mConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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* Z% c' D5 m8 U; j0 a3 \% F h }[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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