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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: ; Y8 H% ]* y& ^* Q7 ~8 X- {8 ~
Let n >1 be an integer
* d; L5 J& l" i0 |- a1 ^. {Basis: (n=2). [( m3 x' }7 J
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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. V) m4 k0 B8 l# \" e) RInduction Hypothesis: Let K >=2 be integers, support that% J2 o! S) u/ H
K^3 – K can by divided by 3.7 J& ^1 G* E+ F* S( P
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
8 x* d* y9 B5 l' \% g8 asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem# _; `. ^6 x9 Z- {2 s0 l
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
( o4 P. d; o8 I2 C& B# c/ d% X% |0 J' P = K^3 + 3K^2 + 2K
8 |, z( I! S0 i = ( K^3 – K) + ( 3K^2 + 3K)
9 \( w4 k# L. o. ~! x* O5 N" N) _. T) q = ( K^3 – K) + 3 ( K^2 + K)1 P$ U9 W+ Z% u
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>07 J4 _# s2 I* N: {0 o8 X
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" m- C$ G7 N. V! G+ [( w V = 3X + 3 ( K^2 + K) |3 u' }4 {3 [* q& U
= 3(X+ K^2 + K) which can be divided by 35 m/ C8 x6 z- N# h/ Q j
3 M+ Q) E2 z# F7 L9 x+ gConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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# n/ r$ G1 G9 w ?[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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