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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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6 k& e1 x+ }4 s4 TProof:
5 h$ n* R' |% R3 m" kLet n >1 be an integer / j: ]+ H+ `. u: Y* O( e- d
Basis: (n=2)
6 A( ]# Y* S' Q0 N9 U 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
5 H* ^7 o: V, b7 F z3 P
7 _. Y$ q! F' `4 R" `" h- E' Q( [Induction Hypothesis: Let K >=2 be integers, support that3 J' G$ i, t1 F" @
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
+ P# y5 c, }# E) J+ S# B) Jsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem8 O8 Z" `1 {6 e5 `! O
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)8 J' }( M# x5 `5 K) B8 S, ^
= K^3 + 3K^2 + 2K
* H$ z+ {( m/ n4 U. |2 ] = ( K^3 – K) + ( 3K^2 + 3K)
) ]+ a& r+ y2 f/ }- c( Q( O" K+ O1 s = ( K^3 – K) + 3 ( K^2 + K)/ U7 B! c' f8 H; @, z% J
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' Q: S& @ ~0 V" B& h( F& D& d) A
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
* A# b5 h7 Y4 w$ u. R5 q = 3X + 3 ( K^2 + K)
! j1 X$ F" a6 |0 g! A = 3(X+ K^2 + K) which can be divided by 39 [6 Y. x+ [1 K! a2 `. [' B p: K6 ]
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.7 T N0 Y$ H* ]; j0 }& `' Y
( H) G: C7 u* \0 `( r. A, R[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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