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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) |' J8 E& a: S. c- F. K, J- H
, E; j- g) J5 O r! ~* n- D4 ]) E3 lProof:
" c8 D+ K: U1 f1 [( v( iLet n >1 be an integer 0 O# z. }" B) K$ L' O3 L) f
Basis: (n=2)
* F& u5 r3 l; k( D' ? i1 m 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' F- _2 y; K% s7 B( T
/ w9 C5 j1 ?0 r8 W5 ?; ~0 mInduction Hypothesis: Let K >=2 be integers, support that
( e* h- L# z# K3 u" s9 N S K^3 – K can by divided by 3.+ w( y) n2 G- B: y/ |$ |0 d
8 D$ y: H5 \: r1 N( {! o, g
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 37 u7 u' r% t$ H, M+ A
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
# G0 W% f" F; _Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)6 C& s& h; m1 r0 l0 N- r
= K^3 + 3K^2 + 2K
& r2 I& @' @, ^2 s = ( K^3 – K) + ( 3K^2 + 3K)
- Y) K% u- ^) ?9 I5 a+ u3 R2 l = ( K^3 – K) + 3 ( K^2 + K)! N" q+ e0 U. w9 h- t
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
' W9 b0 p: G- m+ e; c7 WSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)) V8 N2 X# @+ E' p) o. E
= 3X + 3 ( K^2 + K)
( D d4 i+ F" U2 ^! L4 I = 3(X+ K^2 + K) which can be divided by 33 h$ Y2 `! t$ u6 X: [6 T. X
: q: Y1 l' H6 T. v& _' i& SConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.+ a4 f9 Y$ f u( A$ g/ ?4 m# M
" x5 |# `% M3 j[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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