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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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Proof: : X7 D+ E9 d; Q7 b, T# {. I
Let n >1 be an integer 1 _8 K k" @, c
Basis: (n=2)
, a; L1 Y. p6 i- Q 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3( b. [9 S$ G" q/ y
3 v0 Z# B2 Y/ X5 ~Induction Hypothesis: Let K >=2 be integers, support that
% F+ d7 w, {$ k; {" F' X* u0 R 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
( J/ G) z# ~0 g) L0 M' X/ C9 vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
: D5 g) L; N% H0 iThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)% e- C$ t* X- N9 ^" F; Z+ f
= K^3 + 3K^2 + 2K
, Q# A3 z7 h4 w* j7 A = ( K^3 – K) + ( 3K^2 + 3K)7 B, b; G$ Z. t: r8 M
= ( K^3 – K) + 3 ( K^2 + K)
* W& {- J" C6 B. f* L) Hby 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( v0 m+ ~" k8 Z5 y3 q
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)& [# _2 H% s8 I9 w9 R
= 3X + 3 ( K^2 + K)* f* z% x& {; t7 A: K1 H
= 3(X+ K^2 + K) which can be divided by 3) G2 y/ i% u7 D% W& P
& \' Y9 c4 V9 N# c9 h8 xConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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f1 [& N0 P# ?* e5 [6 b7 C[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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