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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)( ]3 V1 A& V( i9 e
9 g- e8 d% ^& L. N# QProof:
- o+ d8 }& G9 J7 d$ i# f$ gLet n >1 be an integer , I+ d) Y" k, v: I" P, t
Basis: (n=2)( z, j8 D5 x8 P w8 B2 R
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3, T6 C& c6 m5 H4 Q( p
/ @- F. J3 z. b1 U; E, H" w4 nInduction Hypothesis: Let K >=2 be integers, support that
Q! `% S+ L# e2 a- N* C K^3 – K can by divided by 3.
' A- R. k7 u+ q4 f' ~2 I @9 K7 F
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3! S1 ^/ W$ K5 |- k6 t3 x
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem) f/ n- x& B. P7 z5 g
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)+ [' ~2 ?5 z- q ?0 l
= K^3 + 3K^2 + 2K8 e' _) }% x* T/ E
= ( K^3 – K) + ( 3K^2 + 3K)
+ x1 y; o z9 |" C8 P- E, a# o* q = ( K^3 – K) + 3 ( K^2 + K)
+ ]* _$ Z7 ~* C- W, R1 u1 |. ~" rby 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>09 J0 {# F" M3 }6 J! N
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
5 s! m5 Y1 D, {1 y = 3X + 3 ( K^2 + K)
9 a3 J6 t _6 N2 }& F = 3(X+ K^2 + K) which can be divided by 3
6 q: N* V, X# _; T0 F5 M- Q2 S- R- K. b5 q# T- O0 N3 t& |2 u# K
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.( l& _. B. Y8 \+ D2 h! \
[* c1 \& g8 P3 o+ `) D
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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