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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)
9 h0 Q7 h- I2 d# `# k* v* z7 q0 O- D: Y
Proof:
- F: S0 T) h% K& jLet n >1 be an integer & g! ^8 q# M2 X
Basis: (n=2)
0 c* K3 y3 n$ t5 r" h 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 y K5 [# }9 ~2 ~+ i
, e# ~7 H! p+ h; S0 k+ O& C5 v7 h
Induction Hypothesis: Let K >=2 be integers, support that
7 x- ]$ i+ y7 p7 m1 b, `* [ K^3 – K can by divided by 3.
$ y. ?; D1 U+ O# q( F
* Y& E! U6 b4 M) M3 p0 w4 B1 Q5 INow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3" T o' g, K7 ]! t1 n1 H
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem( R K2 r _& Q. i' J. ~: ~
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)9 I2 ^3 S8 G, H2 p' h) s
= K^3 + 3K^2 + 2K+ j6 m1 T, S7 \& {# p1 l* {
= ( K^3 – K) + ( 3K^2 + 3K)
8 U( w. J& k5 c! m7 K g8 M = ( K^3 – K) + 3 ( K^2 + K)# u1 S) b" H1 d' N" b
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; K3 @6 H. Q4 M& p2 }
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ z( M g1 O* k# q = 3X + 3 ( K^2 + K)$ p+ r+ | n$ M8 q7 h
= 3(X+ K^2 + K) which can be divided by 3( ^3 T5 \- |6 [/ e. L
8 u! U/ A+ u7 u6 z: g, P) SConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1., z- [, n8 Y# O' l# d2 L
9 t- b& w: g L1 A5 `4 U[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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