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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)4 [6 n& Q, e' \6 I8 s7 E$ W8 t. U
0 X1 p' S7 ^# m: ?- u- ^( jProof: 9 b Q7 c& O3 I* i: l
Let n >1 be an integer
; ?) \2 N. v4 w* D6 Q0 K5 ?Basis: (n=2)" {# r' I# b% o' X, d% M
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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0 l2 d, Q: k* G6 `7 uInduction Hypothesis: Let K >=2 be integers, support that
" i7 Q# |# _5 N) y# M' X4 b, p K^3 – K can by divided by 3. G2 L. R* y# h: \ g. q
U9 B5 B R7 T! x. z5 _Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
1 r4 N) ?1 _8 ~" O% S4 _since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 x/ v# Z. F8 {- H' mThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)7 T; q7 v! w2 B% i/ P/ b! C$ V
= K^3 + 3K^2 + 2K
, E2 w: X+ g; [8 ^) e \ = ( K^3 – K) + ( 3K^2 + 3K)
9 J2 \; f' Q" g( ` = ( K^3 – K) + 3 ( K^2 + K): N7 U% c$ ~2 ?9 z
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& j# P: S1 F, g1 I2 p0 e# u
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
0 b- r" b1 C3 l4 `/ l) F = 3X + 3 ( K^2 + K)0 k- i! t: X1 Y# K% |5 K
= 3(X+ K^2 + K) which can be divided by 3
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! x3 D7 J4 e6 Q! wConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.9 c3 z# ^8 j( A! e1 N. F* l
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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