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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): {" b/ {) e3 F) p
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Proof: & e( D& b2 f( L/ s( |
Let n >1 be an integer " @) H4 \* v) o" _. a1 N4 @& R
Basis: (n=2)
. j/ |9 d2 [: `1 f+ |1 ` 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3$ w1 y% T; y% @& b
5 i! I' c7 @) h9 s; O; ]' ZInduction Hypothesis: Let K >=2 be integers, support that
+ t+ w0 @. J, j, H" t 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" d- f4 z; l9 l1 _( z& I# e8 C- F; {
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
6 W. {- b! p0 m2 o; z5 d b" `Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)1 _' ?; P$ I& ]& L! }! T0 i6 A) i
= K^3 + 3K^2 + 2K/ K6 G' U) t$ Q
= ( K^3 – K) + ( 3K^2 + 3K)1 a8 ] W& ]6 b0 S6 F
= ( K^3 – K) + 3 ( K^2 + K)
1 t" o- ]2 A6 D% ]; A; O+ Eby 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>06 k7 t5 @* f2 J% Z6 ?$ [. H
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 J1 e# h" e3 B( c! S9 Z = 3X + 3 ( K^2 + K)% H: N9 i2 S% t
= 3(X+ K^2 + K) which can be divided by 3
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, c) L, }! H0 v; Q; v- yConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.- w4 o3 c8 H, b# r& i- ^
! P {% c/ \! _; ?& S; `% f5 M+ B5 q[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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