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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)4 q% H! ^7 Y! v9 f* R3 I0 J
0 h+ v0 {4 s4 ~' `4 g
Proof:
0 ]$ @2 I) k8 RLet n >1 be an integer
( E5 k0 \! n+ n: T" f) pBasis: (n=2)
; n! {4 j4 \: R$ n- G% b 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that2 i- R. X) J( ~( z4 `
K^3 – K can by divided by 3.
: [& C+ I' C$ f- ~3 I }. i* C$ S& }6 T# k3 r
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
7 Y4 G2 `7 R6 i! m+ j' `$ p$ d- {# q1 osince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
! M5 o; z! \0 G: c; }# ZThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)( I" m) U6 p! Z4 ]: c- W) j) _, {
= K^3 + 3K^2 + 2K% V& g: [" H/ d2 U& o
= ( K^3 – K) + ( 3K^2 + 3K)$ B* }) ^3 O$ Q8 M1 u
= ( K^3 – K) + 3 ( K^2 + K)
: o, a3 o, D2 H, qby 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
" E8 G5 f+ @6 [) y: S* T; n2 WSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
( [1 `" k) }0 |- R" N4 F" P2 v = 3X + 3 ( K^2 + K)9 v! Y! j7 H- i0 _, ?8 [0 F; d0 A+ J
= 3(X+ K^2 + K) which can be divided by 3
( p$ Q& L0 E# j; Y0 r/ v& D4 E
+ m0 z% C" w# m9 ?1 n( _Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.& t4 K) M1 ^1 C! E' l5 _
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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