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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)( ]. c8 X- m: V7 L0 F" ]# R$ a
; R9 ^; v' ?6 M. A/ O
Proof: # j: `& R; _) K/ k4 r
Let n >1 be an integer
2 r6 G" H! T( EBasis: (n=2)
6 J) h8 X8 ~8 s& ^* |5 q- P1 K! z 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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: E9 I# J" p4 R+ Q/ BInduction Hypothesis: Let K >=2 be integers, support that8 Z* T" U( U; W E4 D- ?* r
K^3 – K can by divided by 3.4 j5 M: J4 S$ _" u! a) \, K
- c! i1 [4 d; B1 s/ I( M! E3 @Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, e* v" J$ [. b; F4 N* r* vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem& v1 W. U. p* u% X$ H* D- R. o: B
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
a3 ~( Z, c" Y( Y3 b = K^3 + 3K^2 + 2K0 q, ?! N' ~* I; l% O
= ( K^3 – K) + ( 3K^2 + 3K)2 b0 j# o, z0 m' g4 V
= ( K^3 – K) + 3 ( K^2 + K)" V9 i& K+ y5 P+ U) n. ?1 ?0 F
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 B& x |$ r6 t* q/ B3 m/ Q6 _- H
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
3 P8 K H. N6 t$ V = 3X + 3 ( K^2 + K)
. M6 @. F/ F% e8 q2 j& M5 ? = 3(X+ K^2 + K) which can be divided by 34 u& D% L- m o- F
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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% [! O6 G8 E$ S0 u9 h[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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