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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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Proof:
2 D. u+ |+ e, j( |/ _Let n >1 be an integer
& d! c0 R& n- B( W0 D& q/ DBasis: (n=2)
2 H* R; ~4 \/ T' u5 f3 C- z! _ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# m, m& o+ b/ D$ m" s
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Induction Hypothesis: Let K >=2 be integers, support that$ b1 M8 k3 C( }, ]3 k3 m4 M5 t
K^3 – K can by divided by 3.
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" C" @- t. F: Y$ j5 w0 J* T, P6 N( iNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
3 |. W6 g( }' R: zsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
?) e9 b+ p% L, A4 @Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
0 a, |& A3 l5 b = K^3 + 3K^2 + 2K
: ]' W: V: E' i. ~' |7 @ = ( K^3 – K) + ( 3K^2 + 3K)$ u2 n p; y7 } d5 h4 X' r8 R
= ( K^3 – K) + 3 ( K^2 + K)+ H8 _0 c! y! e Z! C
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 i! Y0 M7 h8 y5 A! d6 x1 aSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)* w# W* ]6 z- T( X
= 3X + 3 ( K^2 + K)
; C: j! q, e& M+ M. P = 3(X+ K^2 + K) which can be divided by 3
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; L8 I* x ^/ ]; aConclusion: 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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