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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)
7 F3 ]$ c/ d4 n2 p- D& C/ h1 `: e6 t; a5 s
Proof:
5 T+ ~2 Q3 {+ r0 SLet n >1 be an integer
7 q7 @# E* a7 Z) t5 U" ABasis: (n=2)6 P4 @0 B0 h# v4 e5 C9 h* T
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
+ Y9 M6 h4 X8 X2 B% o9 t% B; v! {2 z" |" ?
Induction Hypothesis: Let K >=2 be integers, support that
; w; s' p+ E8 {" W* f+ \ K^3 – K can by divided by 3.
# b% c3 c1 h; P; C
: c; F! V; N& W# VNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3& N% v& }' z& H$ c# I3 }9 R2 M6 t
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
- Y7 E, p% I! u9 d: s( TThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
: [/ ~7 [) o! L = K^3 + 3K^2 + 2K
5 B k0 [7 ^* b4 i0 z = ( K^3 – K) + ( 3K^2 + 3K)
7 p* ^) B1 i/ ?7 d = ( K^3 – K) + 3 ( K^2 + K) h5 A% `- Z5 ?( A; V
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>08 q: x5 {; a' i- Y7 ?8 [; G
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)+ p7 z: V+ I8 y/ s U# F
= 3X + 3 ( K^2 + K)6 N8 _) I4 s3 s' k
= 3(X+ K^2 + K) which can be divided by 3! y5 b- h2 h2 q' ] b
. \; \) {% I, t7 a1 D
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.+ K' @/ s: P/ ^4 f" n7 Z
% @. t2 \1 N) @& N
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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