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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- q( R% ]4 v1 f
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Proof:
- }7 z1 E! I" i* tLet n >1 be an integer
* f& n2 {2 u. a* i5 k( Y- z# l5 pBasis: (n=2)) b9 u9 X' _9 F
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3/ r% g6 ~) O2 D+ O8 e: K
0 }1 z) g" ~2 @& x8 t& U$ Z1 l& bInduction Hypothesis: Let K >=2 be integers, support that( i$ j* v0 ^/ ]0 h
K^3 – K can by divided by 3., |5 N5 a* p2 C
3 L* F4 e! i6 A& G) MNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3+ D1 b0 }$ q9 J
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
* h5 o0 [' u$ e! s1 K! aThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)+ j: Q. e) ?4 O* C' E# E% \
= K^3 + 3K^2 + 2K
$ I8 V0 S+ g( S: y: d = ( K^3 – K) + ( 3K^2 + 3K)2 m9 {( n& }0 }4 G# e4 v, `) }: P
= ( K^3 – K) + 3 ( K^2 + K); T3 z+ K1 p% V/ s. O* c+ B4 F; Q
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 m" J, p y4 @7 M4 t7 r
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
, v; j) I% f/ ]9 d6 I9 k = 3X + 3 ( K^2 + K)
7 Y" E/ t$ y5 L4 Z4 @3 Z0 s; ]3 f = 3(X+ K^2 + K) which can be divided by 3: A7 r4 E) H) W! h+ J6 e& S! k; e
% f1 ]0 P6 k5 \$ p' F3 j" M) k eConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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; H" _8 C- _" f- R0 G' r4 ][ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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