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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)1 A8 s; g, { ~/ a2 j3 m) c1 N
& [2 r) z: V' R. P2 iProof: # j9 S$ M6 y$ X* X: H4 _3 h
Let n >1 be an integer
! D% q3 D! X! R* SBasis: (n=2)
: D2 ~; {) r, Y% m 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3 l/ H* k/ ? ^# s
0 [6 j/ t2 H( q) f1 mInduction Hypothesis: Let K >=2 be integers, support that
: z! \. M8 O* {3 g7 a% t- a K^3 – K can by divided by 3.
% r% |5 q8 y3 H7 q* f7 X3 H+ A+ h% K+ a. h
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3/ o u9 b/ p, J* f0 z
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem6 J0 U7 l! U- x5 d8 k2 y
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)4 b/ U# C0 c! j# S ~' J& f
= K^3 + 3K^2 + 2K4 {1 |5 S+ e( N) _3 l
= ( K^3 – K) + ( 3K^2 + 3K)
- w, _" s% A: X/ T% F2 f$ F = ( K^3 – K) + 3 ( K^2 + K), {: G! A" ~: P/ B6 _: t: z8 j
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
* }7 {( G- w- j* M% }, kSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 D- T2 P" t) F/ X0 ^4 _0 ^# l = 3X + 3 ( K^2 + K)+ J+ c! C+ O' h1 Z0 o8 R
= 3(X+ K^2 + K) which can be divided by 3& H; Y) D3 `' n3 `5 o" D1 w8 h
- C$ G+ v2 N" q8 m+ hConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
1 `( r/ l3 V$ @1 E0 f( V9 @* T. l1 | i) o! [ S9 C
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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