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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)- a. P: l& l. E
. [ j3 U/ ~/ ]# \; ~' T2 y) \Proof: ( `$ N1 A! {: j# G8 i
Let n >1 be an integer " T F# s- e4 A* [; j5 N" I3 t
Basis: (n=2)8 J `9 p, @+ a! I! @# r' u
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3! |$ _# H2 |- y& r0 {( d
$ B# _, B" b* i0 S( V5 `Induction Hypothesis: Let K >=2 be integers, support that
! l2 P! k4 b: f8 ]' U1 m K^3 – K can by divided by 3.( [. ?% z0 k/ K) h2 F1 {
9 Y% S7 p2 K" L3 x, p1 x. ^
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 30 T7 y2 [/ t( i& Q% N3 u/ |
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem% w8 k8 T7 k. r# a: u" B1 w& w
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)( v) m( T9 i4 ^' N
= K^3 + 3K^2 + 2K5 [, g! k9 ~4 C: m! a, g! I& l
= ( K^3 – K) + ( 3K^2 + 3K)- l$ F* m( Z1 y: X! E0 H/ ]
= ( K^3 – K) + 3 ( K^2 + K)
# f7 s+ n0 F4 A# n3 {5 V+ @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 l( L/ z, }
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) H! u$ k, [: D0 Z) }# w = 3X + 3 ( K^2 + K)# k7 T, ~3 v+ I0 ?6 K0 d" T( l
= 3(X+ K^2 + K) which can be divided by 3
) \: }2 B# ^# T0 |4 q. M* {: ^$ }) ^7 f" w% k |' n& L
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.& b8 g# [4 T$ A( l( Z, H
8 o) ?7 c6 o+ l, _' S8 o0 e' k1 U[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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