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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)
$ R1 a! r0 i$ R/ Y* R" m- ~3 g. N& V' f1 w+ h7 {/ l* X! F
Proof: + z% c& y* k p: u+ H7 |
Let n >1 be an integer
% A5 X2 N6 S1 Q) q1 iBasis: (n=2)
3 m8 G) R' W- C2 J+ z 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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2 N) V2 b8 A, }( ?9 M9 @Induction Hypothesis: Let K >=2 be integers, support that
; _' ~! u6 F! c* K K^3 – K can by divided by 3., ^. }2 K6 D s
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3/ S5 C* i- L' e( b0 ^2 J
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
- Z2 r7 ?; o4 I; ^: H YThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
9 B6 k, q/ s; Y = K^3 + 3K^2 + 2K
v, F t2 A# P- X. R$ |9 l9 t7 W = ( K^3 – K) + ( 3K^2 + 3K)
" c' R) ~# B9 U+ Q& S2 J = ( K^3 – K) + 3 ( K^2 + K)
3 Y9 p0 Z4 H" ]+ Z/ ~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>05 H! J3 X a- s, z+ l: |
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ O9 E/ E( w& J: M) B* h; r = 3X + 3 ( K^2 + K)' w1 e, o) w# d: D. S
= 3(X+ K^2 + K) which can be divided by 3
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D3 d, x" D' c C- t& R5 u, kConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1./ j) q5 n$ \' r: b& W
5 J& }& s4 q- \0 ~7 B4 \0 e[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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