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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): z6 r/ {! U/ E
2 Z! w* u0 y4 g8 J! U* ^Proof:
2 d- U7 j: d! YLet n >1 be an integer
# E- Q& x: [# g1 z6 k6 ?/ iBasis: (n=2)
8 e& m0 d- S4 Z9 e& I) p6 { 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
. k! C2 M X4 U. k
, a" c( K* ~3 j2 XInduction Hypothesis: Let K >=2 be integers, support that
! Q" [' x8 ^' L0 H+ ?5 W0 l K^3 – K can by divided by 3.4 F) i" \# ]8 r1 n% u) T
0 E) w# r+ b3 v4 D
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
* R6 W( h/ Y X7 b6 Psince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem4 W! F$ f+ U7 V
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)- J7 { o A `6 o" k! P
= K^3 + 3K^2 + 2K
: c$ P/ I- O9 Q* y" \2 w) C = ( K^3 – K) + ( 3K^2 + 3K)- B& E# R1 y+ n, B: T* t' I
= ( K^3 – K) + 3 ( K^2 + K)
# T2 i y* h5 R' S1 a% `4 Gby 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
. f# n1 U, W( I8 `So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
5 ^# [$ z8 f; G$ y* j e = 3X + 3 ( K^2 + K)
% x+ [5 h( @8 y7 x4 Y. O = 3(X+ K^2 + K) which can be divided by 3' e& \+ {: ^5 M4 a$ `
* g- ?% r' V6 e6 W/ _
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
9 [$ J1 L5 P$ M( s6 V) B( `; q% G9 Y0 h: `0 D; A G
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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