 鲜花( 53)  鸡蛋( 0)
|
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) t" e x; P- i, D2 j! ]( Z _5 T
. V7 _% z" k5 v1 a% q
Proof: % o. Z/ X: G3 M1 X' d5 [
Let n >1 be an integer 6 W; p' D: T* c
Basis: (n=2)1 T* N2 S. P0 b( h, l5 w) U- U8 z
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
3 w' w, _( @) I8 E& X8 b: a! j/ j- |$ Q8 U
Induction Hypothesis: Let K >=2 be integers, support that X, |6 p2 m- n
K^3 – K can by divided by 3.; x8 k9 F/ J. ` b" w7 G
/ V7 E- A- |+ _9 B7 |- F$ H
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3* j. u9 V) o7 l8 n1 c- [
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem4 G3 s3 O' f# V$ q j
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
! N, g9 c& {0 Y2 `- S = K^3 + 3K^2 + 2K
M1 g! {8 p9 y, A4 l" t = ( K^3 – K) + ( 3K^2 + 3K)
3 `5 ]) L4 x: \: Q8 l$ @0 t = ( K^3 – K) + 3 ( K^2 + K)9 m5 c6 A, g: ~$ x+ `5 B2 B
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' @0 J7 V2 j% B7 @: q
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)4 t/ D8 r* R$ x
= 3X + 3 ( K^2 + K)
; H4 J% A, d S. n5 K: Z = 3(X+ K^2 + K) which can be divided by 3, \ ]: t8 F# N* n4 m% V' q. r
: X4 e: }: `" KConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
8 Z4 h2 j( M2 y' Y
5 h) X3 [# M) \* S1 r2 V, X5 b! L+ R[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|