 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)/ H5 I+ A) P& b. _/ Z- N7 f
: n4 {3 n* S6 F) Y" zProof: & d' q% [) `6 d$ W; _. w$ v
Let n >1 be an integer
* {! W. j/ q$ T. j! j) ^Basis: (n=2)1 u+ u; C0 _$ _3 {# q g
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
* o& R, @- U- z8 @# W" \( X, `1 b' O) D% J1 |
Induction Hypothesis: Let K >=2 be integers, support that
: s) f( F( G" b3 Q5 e* s- u9 b3 `/ D K^3 – K can by divided by 3. o! l* I6 H1 E2 O6 }$ S: k
. |& q3 F4 B5 H: D1 y
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3* C- I" M. _0 Q# `% V1 g2 T7 `
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 h- Q6 |* Z1 K9 w& O, k( }$ p/ i$ ~Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)0 E( M0 G7 O* c% j& K$ K
= K^3 + 3K^2 + 2K% o# Z1 v2 u" _' H2 Z
= ( K^3 – K) + ( 3K^2 + 3K), g$ u \8 q4 t" a( h, y3 q) N
= ( K^3 – K) + 3 ( K^2 + K) b" M" v& k* g9 X" K0 p1 X& B" m
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>04 e5 }$ E- G* |( @- m# [7 T8 t" C
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" T7 J' X: n4 J1 V; y2 I0 I6 ~ = 3X + 3 ( K^2 + K): Z+ ?! i) m3 f
= 3(X+ K^2 + K) which can be divided by 3) q- Q. e' ]: s- G. {! {# |
9 o1 M9 l! H9 k( `; n$ F! dConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
) ]- Q0 e. E, q& _
. Y- l5 Y# _* N" Q+ \9 r* X Q[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|