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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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( Z% F6 H1 R3 N/ H6 g* SProof: " v% w3 F$ R4 }- A) X; n, `
Let n >1 be an integer
$ f* n% R- U$ V; `: OBasis: (n=2)- L& z' R) j* m" U( R
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 35 d6 E% G7 |9 h% R# X/ v
9 d. q* d- V: d- ^; U: f
Induction Hypothesis: Let K >=2 be integers, support that
3 o2 X$ }" q4 O& S6 a, _2 q 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
9 Y1 P% e5 J+ b+ C: rsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
4 _' r/ i0 m- f7 Y) z& y& pThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)5 \/ v+ {* f/ S- ~
= K^3 + 3K^2 + 2K
6 U% F5 F9 t0 [% `: G = ( K^3 – K) + ( 3K^2 + 3K)3 J4 R0 _4 F7 m. W7 H
= ( K^3 – K) + 3 ( K^2 + K)
( K* b) U' P5 t+ k: b( @. xby 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
# t, u8 e' e/ XSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
' T8 }+ A( N; Y7 w3 j7 u = 3X + 3 ( K^2 + K)
/ f* v6 M) ?- f0 v- ]$ M* o = 3(X+ K^2 + K) which can be divided by 30 k( o, Q7 }- P, u; g$ V
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.5 Z* `0 E# S# r
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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