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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)
% H; h+ O$ @3 }3 l6 e5 ^# Q/ E
* L( s6 ^8 M. lProof:
+ b. [+ B+ Y- Z' W' U, N! D1 kLet n >1 be an integer * H- S; c4 _0 t- [: {9 e/ R
Basis: (n=2)- d" v* I" ~2 c4 b8 ^
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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. n w, @% O& x$ x$ `& I4 H& WInduction Hypothesis: Let K >=2 be integers, support that
8 r* l" b1 V# t' @ K^3 – K can by divided by 3.( _- Z3 r! u% g3 u! s1 Z
3 c* U, X& q4 ]! oNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
# K1 {# p' r7 O' S0 J1 L# psince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem! D0 Z; J7 f$ e' X4 @+ v
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, R6 C- {: R T2 U = K^3 + 3K^2 + 2K# s* V7 U) O$ K1 L$ y( P* d
= ( K^3 – K) + ( 3K^2 + 3K)& `" o* E" b1 D) z1 r
= ( K^3 – K) + 3 ( K^2 + K)
( O9 B% ?) n- {, U1 ^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
8 F" X1 c7 ~% z$ vSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
& t+ H/ g; L: v4 C" @ = 3X + 3 ( K^2 + K)! K. v( x1 W7 N4 i2 ~
= 3(X+ K^2 + K) which can be divided by 3
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( u% \$ k' ?6 R+ l$ y' pConclusion: 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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