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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)- b2 K0 u' \& k/ I! t0 b
% X7 R' K$ \1 ^! |8 O( fProof:
. {& v/ G9 C% u6 c: l! C' [" ELet n >1 be an integer % C2 U( q* C/ s( o( S
Basis: (n=2)9 l; K2 K, [; l k; Z8 @- {! e
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3 a ?0 M( ?1 r, ~% c F7 h! l
# G# V. {/ I/ U* _ `" J c
Induction Hypothesis: Let K >=2 be integers, support that
2 a9 @! _% n# d K^3 – K can by divided by 3.
5 y8 u1 [, T: n7 o: E" e8 R5 C* @4 m, F; V
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
/ u0 w2 H6 B3 [( b1 Y: p, Q/ x. S1 Asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
% o2 @ k5 ~+ u: bThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)0 E3 H+ a0 |; P: w! J
= K^3 + 3K^2 + 2K2 X6 W' ?/ D! p4 ^) P
= ( K^3 – K) + ( 3K^2 + 3K)
1 @- B9 z g/ L' }6 K) Z; r& b4 L = ( K^3 – K) + 3 ( K^2 + K)
9 i3 d" L8 m' e$ k0 Cby 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
" [+ A# Z7 S$ m- P% ^So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)+ ~+ g8 [$ W+ z, b+ z( v6 [4 |+ N8 Y
= 3X + 3 ( K^2 + K)
# J/ W; P+ p+ M# z r# `/ r = 3(X+ K^2 + K) which can be divided by 36 Z" k+ Q P5 {1 c7 i' ]
+ r4 q4 a4 R8 n. Q: oConclusion: 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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