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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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- _) @+ V# v- f U$ s! bProof:
% w: B/ s# {0 N/ T$ g1 ULet n >1 be an integer % g( A3 t2 C& C/ m2 Y
Basis: (n=2)
& ~! C( B; P+ s' P; J: y 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 35 A H( J; n3 I/ T3 c. ~
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Induction Hypothesis: Let K >=2 be integers, support that: S- W! h' D2 M$ v
K^3 – K can by divided by 3.
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) x- \# \/ o' w; F: b ^Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3. _& O' e! c5 F' l! [
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem0 |' A) o5 W3 w3 y% `% C
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ b. S/ v9 Q$ L = K^3 + 3K^2 + 2K
" S( R6 ]$ o/ Q% P = ( K^3 – K) + ( 3K^2 + 3K), x/ [& c- q' Z- Y& @! l5 p+ I
= ( K^3 – K) + 3 ( K^2 + K)
, _9 `3 k9 p* j: wby 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
; X' r5 d% K; j8 H/ r& gSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
; G) f- c( c" e = 3X + 3 ( K^2 + K)3 g. i1 k2 U/ ]
= 3(X+ K^2 + K) which can be divided by 3- W, c5 ?- z! }1 n. j2 _
, W: D8 g% i3 f5 fConclusion: 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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