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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)
( V. d7 J; o2 E2 B
& |' ^8 z" e& {+ p& ^4 `Proof: 5 v9 x8 T; y" O5 p; S# t
Let n >1 be an integer
2 [" e7 C* D zBasis: (n=2)
8 A. V/ `# ^2 h W; w- s 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 37 @$ V6 Q: n; I- g( U' ~
- ~3 j: j% S' mInduction Hypothesis: Let K >=2 be integers, support that
9 h; j9 G' U5 ~, l+ p K^3 – K can by divided by 3.2 a6 w) P$ D r
& K2 k5 G- D( k( b4 |Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
- j9 @ b: c1 }! {; ~since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem* t, e2 y! d0 L
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
9 p$ ]. P) C: S& N4 g( ]! K3 O- q = K^3 + 3K^2 + 2K
7 Y- z* Y# l7 J- E2 d+ `0 Y = ( K^3 – K) + ( 3K^2 + 3K)
! R: k) A2 j& I- `! k S0 q = ( K^3 – K) + 3 ( K^2 + K)( `1 T" z1 {. P& F
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$ B6 C8 X% R2 a9 g4 L# A
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ s; w0 I1 L$ A- N = 3X + 3 ( K^2 + K)9 \1 K! M4 ^: \) ]
= 3(X+ K^2 + K) which can be divided by 3
* H, @; N2 ~7 V- |! M# n3 |( G% n% @1 ^9 D7 t
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.1 \, M; S: h4 q2 _4 C( \" q9 r
1 l# N9 O0 d8 `/ Q[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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