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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)* c) h* d" a* g
) _* x% E/ F/ z, R# B! aProof: o; W+ c0 b! G1 r/ E
Let n >1 be an integer 2 n$ j+ s2 N3 q/ F) p2 g
Basis: (n=2)2 Z! i% f$ K* B k# b! T
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
' C0 R+ r3 C9 K1 L1 ^6 S4 M! j, q2 ^0 ^+ I
Induction Hypothesis: Let K >=2 be integers, support that
& m) S7 T7 q# F% ]8 y& r' f4 } K^3 – K can by divided by 3.
4 a0 e6 H( t6 ?( H' c9 E# B/ u( t! a: f5 b$ k
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3) z' x N5 C) k+ h, M
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
; r0 ?4 G9 q/ b! ^: \Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)4 Q- W2 {9 C8 }! ]: |
= K^3 + 3K^2 + 2K
/ R, r" [9 j- _6 M$ e" @% n; C$ ` = ( K^3 – K) + ( 3K^2 + 3K)
* j3 c$ |( H! A8 d* n = ( K^3 – K) + 3 ( K^2 + K)
+ T. c9 O6 g' a& V3 d1 wby 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* ]- g( v! T4 X% d3 o
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
! R: s3 [! }) G9 n; w, f) H# J = 3X + 3 ( K^2 + K)
! {2 L/ Z v1 O = 3(X+ K^2 + K) which can be divided by 35 S: N4 v" R$ t$ G. s4 R+ Y' u9 J
6 [/ |: ~( y& d! Z! x$ C* _
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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+ b9 U. l9 v- K6 g5 i$ b! h, X[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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