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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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. }' F2 O9 H9 b8 K4 f3 F& MProof: b. U4 b3 _( M A5 d4 U
Let n >1 be an integer ) c# X; B8 g. o# D4 v8 `
Basis: (n=2)
& C7 C- M0 R" Y0 n9 ` 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3( d$ o/ ?& X& X: D* @
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Induction Hypothesis: Let K >=2 be integers, support that- |. d @$ G/ b7 v n$ ^! O
K^3 – K can by divided by 3.8 d+ F4 N2 j0 ?5 y7 [+ i
4 G& g7 S' a1 [% n) v7 n; kNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
* D2 z9 z- `+ j! z U7 Wsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
' E) s6 P2 o/ u4 _4 pThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
8 ^& b1 P$ \( |7 }9 u2 @% W/ p = K^3 + 3K^2 + 2K& d3 [, n+ L4 q4 @
= ( K^3 – K) + ( 3K^2 + 3K)
3 q. k' b& H' N/ ` = ( K^3 – K) + 3 ( K^2 + K)9 c! Z* u1 D# b, P5 K0 k
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
# n6 v2 b: N$ J! c3 m2 f) W' Q! uSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ R1 j3 r6 b7 x! V = 3X + 3 ( K^2 + K)
) \/ G o) v" E" S* C$ O0 P = 3(X+ K^2 + K) which can be divided by 3( g0 i# B& m4 W4 Z- R
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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& V- L! u+ g$ R i$ G, d: w+ j, J0 }[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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