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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)) e2 S* J- Y/ m( ?6 b, ]
+ V9 L+ E/ K9 ` s. D$ Y! lProof:
; Q. U& b1 }# d4 C0 I( e/ O. U' qLet n >1 be an integer 9 p9 q" S* U7 p. @9 `" d$ ~
Basis: (n=2)
( `! _) U9 g( D 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
9 ?5 F, ^: Y$ j
, U* @7 t$ K* t! G4 v! jInduction Hypothesis: Let K >=2 be integers, support that
4 ^' N# e/ w6 ^1 T! [" ^ K^3 – K can by divided by 3.
; V+ z- \$ M8 v0 l: X* J0 s' }/ b- K4 l' U2 h
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
7 l# H% |- A. a2 O% D- Wsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, @2 @0 w3 D5 A, M1 x8 U2 SThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)$ _- s/ H/ `8 P
= K^3 + 3K^2 + 2K, }9 v- I- r/ b: h
= ( K^3 – K) + ( 3K^2 + 3K)
4 E1 [$ p% V+ b% c = ( K^3 – K) + 3 ( K^2 + K)* G, t3 k! _' o' I
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
3 B; X% J: |- J# ~# _' FSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
! E% \! b6 g+ S" M- w) L = 3X + 3 ( K^2 + K)
1 C$ h: E- n" C = 3(X+ K^2 + K) which can be divided by 3
# G! k6 f: r- j& i- \
# R9 B6 E, Z3 d7 ?Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
# E6 `& N5 _, S% A, }) ]+ m5 @0 e9 _) F7 c% z, t. M* `
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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