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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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$ j- V: {' H* P7 l- y- d/ R7 RProof:
# D$ e) o: i M" Z$ GLet n >1 be an integer : Q+ T6 e6 \0 B# @7 a Q
Basis: (n=2)- ^$ _' t" b1 v4 J6 Y$ k; `
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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/ i% _ M; U, S8 @4 x: f+ fInduction Hypothesis: Let K >=2 be integers, support that
7 m# u5 W; K/ w5 V( d K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, U- C" p1 \& I- |2 P- `% R/ s, H! gsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem# `0 z8 \1 i& k% l& U2 Q @) w
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1); D' v6 f5 [1 j; |2 e7 h3 s
= K^3 + 3K^2 + 2K
! y& b4 k+ T: I( m' j6 j = ( K^3 – K) + ( 3K^2 + 3K)
8 e6 a N4 L" p& ]+ ~1 q4 J = ( K^3 – K) + 3 ( K^2 + K)7 z6 `4 `/ S+ Z# T) s5 o
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' Z5 k/ ~: x) F
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
& O5 l* H+ O- ~3 A7 L, R = 3X + 3 ( K^2 + K)) z8 W, |3 G: m$ U! L
= 3(X+ K^2 + K) which can be divided by 3' R0 l% N1 ^( |6 `8 @8 b9 d
2 H. l% L: h) n! g# K# ?" l) DConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.) T9 o ]! F0 i$ \9 q1 U
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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