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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): m J) b) e; M* y8 m: y9 u( s* \
3 R* J8 L' u$ KProof: & g; M) s0 r) S8 i7 c; Z
Let n >1 be an integer ' A: T- j; T8 v; a( n9 U" }9 H, s
Basis: (n=2)4 k4 l/ K% Z3 P' r1 }+ D
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
- k5 \% h% {! z
7 h% T2 u! H7 m3 MInduction Hypothesis: Let K >=2 be integers, support that
2 b: r, a- U5 M8 B7 @. p9 M, N' T3 W K^3 – K can by divided by 3.
0 w' K8 x" ^- Z0 h4 j2 H6 H; l# @8 F. {3 t0 @$ `& ~: ], g
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
. G! s# F. z% o9 o: _since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
0 i! n: Q. W* ?5 \- `2 BThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
6 k" o' a1 R9 D# D: J5 T4 n3 o2 o9 d = K^3 + 3K^2 + 2K( R7 G/ i- u! h) |- o% l! l- s
= ( K^3 – K) + ( 3K^2 + 3K)( i: m6 z6 Z$ r5 S- f4 \" k- D& M
= ( K^3 – K) + 3 ( K^2 + K)- X/ c9 x& C& M
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+ v; H; k {9 z' |* ~1 u- ]2 }9 J
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
# l) F% l) d; M$ ^- T2 V = 3X + 3 ( K^2 + K); S# Q' K# {# x* a
= 3(X+ K^2 + K) which can be divided by 3
; }0 @: t) w& \# K3 G/ |
; F( A" J7 x Y$ o# v' w3 @0 oConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.& ?1 Q# L& D7 E
1 E3 F, |( }, D, d[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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