 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)" E: t; s8 F" b6 U8 k, N
8 s+ B' s( q# O2 X2 r0 QProof:
' C. Z+ P" c1 wLet n >1 be an integer 0 L: \ N$ B* k8 K% ~
Basis: (n=2)( l+ w; R6 P8 S( h, Z/ m
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3: O; i/ K( O" j
Z+ _3 x) I( v6 a+ Z/ F1 qInduction Hypothesis: Let K >=2 be integers, support that$ i3 z, D e( w6 o9 [
K^3 – K can by divided by 3.* |0 y# H* d( f7 @* n) R( M0 g
' I; o5 C, \& s- w" t. gNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3" |+ r( Z4 L! `
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem/ G3 O. R% p r
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
( e) \( U' S, a8 ]7 A = K^3 + 3K^2 + 2K
* u4 X" F& D+ |# R* Q = ( K^3 – K) + ( 3K^2 + 3K)/ f# k9 n) u0 w- S$ ~) f9 w
= ( K^3 – K) + 3 ( K^2 + K)* F3 X& I3 Y$ ^! y. H
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- D- U( v* M" }. n) }
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)' Y& ], G" K/ _- e
= 3X + 3 ( K^2 + K)) H9 q$ r: t L. V
= 3(X+ K^2 + K) which can be divided by 33 Z3 w7 ~9 P3 p$ b* D
' [. @ `3 k) E9 sConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
) M* I/ ?' _$ | o0 c8 q8 w" p4 `
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|