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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: 1 X: o* T1 V0 B: G8 C9 i, x. m
Let n >1 be an integer , }" t* l) P& `
Basis: (n=2): x/ ?* u+ Z: r2 G M6 d( P7 d; t
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
" G+ I! W t& q( L3 h8 o: k' k) `1 b2 R9 G
Induction Hypothesis: Let K >=2 be integers, support that5 W ^) @$ ^, P' o. K+ z
K^3 – K can by divided by 3.
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/ K5 i" c4 i. o8 S6 qNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
5 a$ s' Q7 g- B$ s7 g2 [since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem9 C- U- c: a* A3 {! x: R
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)$ q# C% d, J- T4 l
= K^3 + 3K^2 + 2K
$ V+ o6 ]0 ?9 A! p4 I5 L1 O) K = ( K^3 – K) + ( 3K^2 + 3K)
: Q% Z: b/ g8 N0 [# N, C, o) I/ U = ( K^3 – K) + 3 ( K^2 + K)
" l$ R, l) Q9 @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>06 x8 T* Z' e$ j7 a" [2 Z$ ]
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)! g- W+ O; t" I% i$ D* w
= 3X + 3 ( K^2 + K)
/ ?' C8 [) Z. I$ K, L& h = 3(X+ K^2 + K) which can be divided by 31 e4 @# \' j( |+ X
5 T! l" O: Y1 W" eConclusion: 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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