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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:
, n& K- H! J y9 ~6 |( jLet n >1 be an integer
: P: { C/ ~2 V, d3 ABasis: (n=2)
3 g9 i) E- ?, x4 b# V; Y# y 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3$ F: _. e2 F Q5 R( G$ h: M: O
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Induction Hypothesis: Let K >=2 be integers, support that
+ e1 L/ j' \% ^0 d( C' d 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
~: E. E# F2 o J' Lsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem+ Q, p; N" |% c$ U' }$ Q' ?
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)9 r- y4 B ]5 T
= K^3 + 3K^2 + 2K$ |3 f) `/ ?2 F1 a1 f5 A
= ( K^3 – K) + ( 3K^2 + 3K)
3 \& z" T) p9 ~5 C9 l. J% |! V# q = ( K^3 – K) + 3 ( K^2 + K). e/ B! g% m3 r) J1 |% P* I5 x
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
" _) P( f: h8 O- NSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) [" G1 p& n+ v; ?2 ~2 @3 I. g: r* B- L = 3X + 3 ( K^2 + K)
2 q+ J* {9 U& h" s = 3(X+ K^2 + K) which can be divided by 3+ d# Y% [0 B* Z3 T; y. s
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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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7 Z% \/ d% K% W5 Y# f[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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