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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)
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* }, ?6 i+ t' H3 n; GProof:
/ h. \+ L" h, kLet n >1 be an integer 9 `5 w( i$ a& t% V( i% ~
Basis: (n=2)
0 F8 h( t6 L# f: ^" D4 F 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3+ y- ?: @8 a7 z( c/ {0 b
' T& S# }& \# ]1 Q( wInduction Hypothesis: Let K >=2 be integers, support that
, i5 t. q3 A6 O- \/ N Y K^3 – K can by divided by 3.
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, o$ w/ U/ P v5 }7 ~" |0 F+ iNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
" F8 c6 x5 f {9 Gsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
- |* f: z4 h4 o1 G v) {Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)% [( w+ s. R8 c
= K^3 + 3K^2 + 2K
! K3 l0 s# s1 ]2 d- d3 Z* q& ? = ( K^3 – K) + ( 3K^2 + 3K)
* R4 ^& T; E! |! p1 l9 ]' t4 x = ( K^3 – K) + 3 ( K^2 + K) S9 Q+ v. N% C, G: Y
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+ }6 }% N: } l1 S$ [
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)) w$ i2 `/ m+ [$ ^/ H
= 3X + 3 ( K^2 + K)
$ [- {4 ^0 b6 M6 a! t4 g+ r = 3(X+ K^2 + K) which can be divided by 3+ l* ?3 G2 L" E" F1 C' ^
) p, d+ d2 ]1 ~6 C. h- v9 kConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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# b* l3 e9 i# N: n l" P[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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