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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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4 f4 h, Z4 ~7 N* WProof:
\) \( Q2 T0 ]( e _1 `Let n >1 be an integer |+ k! U/ u" s( h* F) \7 A' u3 l
Basis: (n=2)( a- [! B+ |9 X D6 `$ E4 u' u
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3( m1 L% h& [! o5 M( \
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Induction Hypothesis: Let K >=2 be integers, support that
; ?- b8 b) B& v6 S2 g2 F* a+ o/ r K^3 – K can by divided by 3.4 _# j8 H' P5 @* L
# q Y. U+ m" m% YNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
4 s8 l8 B: {! d. a, M! ]2 ssince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem6 W% w0 D' J. x8 W0 {% J
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
& F( y! _7 L% J" o' ?. X" F = K^3 + 3K^2 + 2K9 S B1 B/ E* u# y7 I
= ( K^3 – K) + ( 3K^2 + 3K)3 [& t9 N2 i# w8 b
= ( K^3 – K) + 3 ( K^2 + K)/ G3 A) m, k" d9 l1 z, y4 Z
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
& [9 t; r+ H( E% @, fSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
( j5 t1 Q; K( O = 3X + 3 ( K^2 + K)& I6 k9 l! b8 Y2 o: Q8 J
= 3(X+ K^2 + K) which can be divided by 39 N0 Q' [8 V; s/ x3 G6 a. N
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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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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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