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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)+ D) x6 n: C6 ~( o8 d9 l E
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Proof:
2 K. i9 R. @/ }5 o9 L# `Let n >1 be an integer
: t+ s5 P# E# f7 J7 y0 t% YBasis: (n=2)2 M) F J+ ]' M; J
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3% g `' I4 n1 c3 n5 m9 _& Z. Y+ I' U
, G! J, r1 V4 | Q7 [: [% n! J; pInduction Hypothesis: Let K >=2 be integers, support that
! [5 Z# D7 @- Z/ Z$ K# t* Q K^3 – K can by divided by 3.$ X* p2 |; d" d* S4 l$ o
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3! S4 n$ z1 u1 N. b
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
$ I! `. E& `9 ?Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. Z9 {0 H1 a" j( E = K^3 + 3K^2 + 2K; J( D: J6 i/ z
= ( K^3 – K) + ( 3K^2 + 3K)
9 A) j( I$ G) r# a9 r& Z = ( K^3 – K) + 3 ( K^2 + K)
3 {/ H# ~) N1 w$ T/ xby 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) w8 R4 f4 K6 ^# j% |2 @
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" p' u4 x2 W9 s7 ]9 q = 3X + 3 ( K^2 + K)2 P4 E+ x9 R0 s# r5 K. U2 ~' b: y
= 3(X+ K^2 + K) which can be divided by 3- q" P& G# Q) \6 ?; I
( g* r+ T n9 q$ ^) JConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.' U+ P7 [" U, V1 _3 u1 q- b9 C
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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