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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( Y, p& N- e% {
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Proof:
" ^, H) @" V% g( aLet n >1 be an integer
# \$ R9 O4 x+ M. P9 X* V, c" B" BBasis: (n=2)2 b+ C7 L9 s5 ^* X' @
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' {! A6 `, t7 b$ a% M
1 k- Y0 J' j9 M4 S. V1 DInduction Hypothesis: Let K >=2 be integers, support that
, |6 I& r6 ~" x4 i0 W3 t& @- V. ^ 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# T5 k4 _/ I7 D- V9 l& ^; L! P
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
' D; t, Q' a+ j, \+ V" xThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1) X0 D' r9 s+ x: ]! Y9 ?
= K^3 + 3K^2 + 2K
6 M9 W( g" Q& W# } = ( K^3 – K) + ( 3K^2 + 3K)! }; [" l* D% N5 t# u
= ( K^3 – K) + 3 ( K^2 + K)/ _$ C. K9 G) m. X5 t+ l
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
1 O7 S5 f, {5 Z" Q: Y$ j8 g9 ^So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)# }5 w \! C8 ]- H; _0 T
= 3X + 3 ( K^2 + K)3 g# z8 B7 W* N( _; H* J
= 3(X+ K^2 + K) which can be divided by 3* \8 J* H* R, A# Q2 i, K4 }
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.5 T! m7 O, s; Q3 N* Y8 D2 f
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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