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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)+ y. |8 C4 @3 B' O" q
; ]2 z8 U* n+ h! MProof:
$ G8 M1 `' |1 S$ l1 g MLet n >1 be an integer
6 v$ `$ | I0 O3 B! V; N0 h( kBasis: (n=2)
: `$ N7 `$ B' P$ W6 U: z 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 z N$ Y" ~! \9 a
+ f) P0 x( v1 l
Induction Hypothesis: Let K >=2 be integers, support that7 P* I& N' R0 [5 M4 M/ T
K^3 – K can by divided by 3.4 A8 { y& G& l
; n0 r; `% Q0 L8 n" i- LNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ Y" R2 x D& G/ H" `8 i* Ysince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem4 s, S8 j' O' I
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1) N% J2 R5 Q" ~8 O" B+ M% P
= K^3 + 3K^2 + 2K
* I, t$ O- D5 g8 {: P% H = ( K^3 – K) + ( 3K^2 + 3K)
, C B. p$ L8 V = ( K^3 – K) + 3 ( K^2 + K)! [* G9 m2 ^3 W( r8 A
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 S$ w+ a, ^- F, ~; @1 @9 U% aSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
2 E4 ^! q1 s% O7 ~ = 3X + 3 ( K^2 + K)1 {$ x( h* f; D. J8 G
= 3(X+ K^2 + K) which can be divided by 3: a' C! B2 u9 s" \
. W7 E" }& n8 n3 DConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1. n5 Q) L% N G. y. Q- F
. k8 C5 B1 m( P" O$ c( Y' q6 y" {[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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