 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)& Z. g& ? O! R$ s) m. [4 I9 f
& \5 Q4 z' ~( w
Proof:
* }5 X9 x1 ?6 w' D$ kLet n >1 be an integer 5 |# s, R5 T n$ ~
Basis: (n=2)- [' e9 o) d' n* c
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 30 V. {1 G% ^0 }- {
2 l( }. X. o# @- {* a3 M
Induction Hypothesis: Let K >=2 be integers, support that
4 B; A! b0 G4 u' [ K^3 – K can by divided by 3.3 F0 L* Y. S z% v( p" j- ~; _
: h7 w5 i: M8 L# b; yNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 33 @) z. w6 A& Z5 f. f. V
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
; B/ _5 P$ k W" r8 w/ {Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
$ B' l5 {; {) O: e* N$ T = K^3 + 3K^2 + 2K
+ p( v3 S7 v& o = ( K^3 – K) + ( 3K^2 + 3K)
; L+ N+ K& }( n* d) U; a = ( K^3 – K) + 3 ( K^2 + K): u2 {3 @9 A) P% 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>08 `- b& Z' b. j
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
0 i& w2 N) Z' {5 a9 z& k = 3X + 3 ( K^2 + K)/ [# C; Q. _( Z
= 3(X+ K^2 + K) which can be divided by 3
3 x k# P6 g, c' h4 I% L' V
2 W0 y; @2 W& r3 z' L( V$ N$ oConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.: r) P* b2 q4 Q( B
! z/ W6 G3 K7 A t4 ~% x4 L# a9 o[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|