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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)) B, x7 W+ D9 l& _3 J
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Proof:
4 U8 e+ S+ w. B$ W0 K T) _Let n >1 be an integer % S2 X3 `( X2 y% @, E
Basis: (n=2)
$ g$ s8 V6 t" m# M1 @: i 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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* D8 R& ^( s3 {# K2 h# r) DInduction Hypothesis: Let K >=2 be integers, support that
& X+ w1 I4 p u: q3 o* n K^3 – K can by divided by 3.
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5 x3 ]4 G* n, {" `' \Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
' Z5 J, R i: g# B) u8 Tsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
; y1 C2 F/ ?8 K: bThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
8 T1 ]/ d' A5 R( w# @& k8 r1 q- ]- Z = K^3 + 3K^2 + 2K, H' ^& h6 d: S* j7 G
= ( K^3 – K) + ( 3K^2 + 3K)
2 S% x1 x9 c+ E& u L5 O% o = ( K^3 – K) + 3 ( K^2 + K)1 N* _+ B) X1 `# g8 K0 M
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
: f2 K0 O4 \4 i; S% f* Y( ^So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" e( \0 B+ K( O& i f2 J = 3X + 3 ( K^2 + K)
8 z- Y1 J* S2 D/ g: h = 3(X+ K^2 + K) which can be divided by 3
% ]" H0 y& i% c/ ]6 w' W. x
: p6 O4 C& V! @- YConclusion: 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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