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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): @) i' N F5 {# W& \+ B$ q
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Proof:
: ~# g! L4 Q! ]# S6 ^! E6 kLet n >1 be an integer $ K( V1 v/ D% e8 R3 P: u
Basis: (n=2); b( y1 B9 B( \! w3 G6 S
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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/ w, ~4 K4 `5 S, e! x: n8 I% k/ }$ B. M+ RInduction Hypothesis: Let K >=2 be integers, support that
8 k5 m1 ?+ `9 {' [2 p K^3 – K can by divided by 3.2 i2 C: d0 I7 @# o; h& J9 {2 Q
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3/ \- q- ]6 @; r$ q
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem+ ^* i8 x* @; b& l; P9 o: E6 b7 h. F
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ c2 M" E* I5 Q' o" v = K^3 + 3K^2 + 2K
~# g2 J( E3 N! H0 x = ( K^3 – K) + ( 3K^2 + 3K)
: x; ~* l; h# E1 | = ( K^3 – K) + 3 ( K^2 + K)6 r9 h, C5 Z1 G2 o- i* U% d3 ~# ?8 g
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 s* b: [, Y7 Y" I8 Q% f
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ a* n) A3 {6 V# h1 {0 t7 j = 3X + 3 ( K^2 + K)
8 x1 ` f3 _( B5 ~) e5 F = 3(X+ K^2 + K) which can be divided by 3
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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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5 U$ h4 \. k- t6 t* F; R- j5 ][ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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