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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)* ~: @5 y/ V, ~5 e( j [9 b
" k7 z2 i+ R5 r# BProof:
! A$ _+ f' g- ?) p9 Z6 ~$ H5 q# \1 ILet n >1 be an integer 3 @& }& t2 L# ~* U& z
Basis: (n=2) D% z3 K* H2 Q5 B: ?7 C0 L
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' D. C( b( \4 J3 ]4 P; M
# O* B# _3 t( I) J6 Q& zInduction Hypothesis: Let K >=2 be integers, support that: Z( x, D1 m/ m, B+ E
K^3 – K can by divided by 3.
, u. p0 ^3 N! z* M; r/ R( K; m8 d- C2 [; w% R7 B+ Z0 f
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
$ z+ S4 ], ^5 P: G7 e g: Msince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem3 n6 U( {. Z% E
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
# i- G) H% d" g/ w" E = K^3 + 3K^2 + 2K. y" Z% v( p, z* G" l- X
= ( K^3 – K) + ( 3K^2 + 3K)" m# _, y, D; G$ l; a5 L
= ( K^3 – K) + 3 ( K^2 + K)
1 B; w0 A: v% j+ o5 \3 R$ z1 d4 Dby 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
4 d) c5 h8 W/ r- D! L# ySo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)$ W+ g9 T! [' E
= 3X + 3 ( K^2 + K)
2 [0 ?9 W. l1 z$ ?6 Y% O4 w/ P9 | = 3(X+ K^2 + K) which can be divided by 3
; T/ a5 M+ H! y6 v, ~
/ `8 A8 B, l# V" c/ ` TConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.' i i$ m" |+ x% H# F9 z1 N' |7 e% |
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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