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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)$ {$ Y- q% O0 @ Z- @
4 L) D- v" j2 E% mProof: 4 O8 x0 y% O% {- X: U- B, n B
Let n >1 be an integer
, I3 N( ~1 ^8 U/ k8 aBasis: (n=2)
6 y" j, I& U. D" v" w4 P+ o& S 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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' q; X; e& X* r; I# c) `Induction Hypothesis: Let K >=2 be integers, support that
7 R* H; G3 e8 g9 m, { 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; v7 `1 z0 [9 J, S3 s5 `
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem4 u& M% f$ a$ B- p: B4 v0 `. C
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
! h6 F8 d- ?# X& M0 Z6 }: q* H, w = K^3 + 3K^2 + 2K" l6 N- ^* r+ T
= ( K^3 – K) + ( 3K^2 + 3K)) G, |5 T2 k. A; N
= ( K^3 – K) + 3 ( K^2 + K)$ _, o: @3 F) o
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 W W I7 E6 ~, }' S' g
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 _4 c6 U3 F4 y( t' q+ K = 3X + 3 ( K^2 + K)
- y3 i& T/ y3 G, N1 S = 3(X+ K^2 + K) which can be divided by 3% T' }7 `3 C3 D6 S
2 A3 R8 B# j' v$ `Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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