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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): a, B7 N, Y0 r3 w9 j$ e
- ~& @3 o+ y0 A- W9 O3 V$ E! EProof:
) L/ F' Q5 u! }. h! \Let n >1 be an integer 4 l/ B" t: w% G# t% S
Basis: (n=2)# O; U6 `* {* `& a
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
8 i$ q* ~! D3 x6 E 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
; ~ w0 t8 x& B, T9 csince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem- M( z9 Q; v2 a b& V
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' v* I# P2 |4 ^& P. p; u
= K^3 + 3K^2 + 2K4 `% z% M' d* x& b
= ( K^3 – K) + ( 3K^2 + 3K)
$ e/ g+ @; D, r/ L1 w! p = ( K^3 – K) + 3 ( K^2 + K)& U* @5 b1 }: C. t: V5 s
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, a6 Y( ]( _8 m5 ?- g+ ~) h
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)4 t( [% S+ V; x4 ?: z M+ S6 |
= 3X + 3 ( K^2 + K)
) T; D" Z8 v. r6 n5 N5 A. i = 3(X+ K^2 + K) which can be divided by 3; ?, [ u/ ~- S. O3 \
0 Y0 {% A) H6 }1 I& X( VConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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0 F' D6 p# o3 o) r[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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