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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): g& f& ? P% W: S
0 j5 i) A; ^5 F f Z9 |. n+ ^/ OProof: 1 L4 h+ X% A; q [
Let n >1 be an integer
! W, O/ f: t" v f( B# @+ L2 L: _Basis: (n=2): J2 Q. K% f) K' W2 l. ?3 J, x
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3& F0 q! f; F" {$ Z# \' j/ ?
4 a- f! G O* e2 a3 R F# B) a3 X. I
Induction Hypothesis: Let K >=2 be integers, support that
: U" j# [' \% a4 l( y$ B K^3 – K can by divided by 3.
) I- ~) p: B; I9 {0 F, G& k3 `2 \9 p, E6 W& l
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3) {7 d8 v3 C, A: b; ]+ f1 [6 i. `
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
V+ c7 `2 ~% a% m/ z* x" hThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' R4 h$ X3 w! n4 y9 E
= K^3 + 3K^2 + 2K
9 q- l& @* |6 |* I = ( K^3 – K) + ( 3K^2 + 3K)# J: i) K& B2 H3 w
= ( K^3 – K) + 3 ( K^2 + K)# Y, v8 ?5 r$ R
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>00 t0 o# z) Z: J+ [( ]2 v! @( V; |
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ p: @$ m N' ~0 e = 3X + 3 ( K^2 + K)
3 n# t# [* c; {9 C: f = 3(X+ K^2 + K) which can be divided by 3
/ J# a9 c$ k1 x: Q$ w
; H: \$ l9 e1 T9 Q; e; [% p! {Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1./ W. q0 x3 k: A! D/ n4 X7 o- `
s: [6 @0 |2 X% i, u* c[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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