 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); M3 w6 t; ]) W. @0 L4 T
# E) U# M9 E1 I# p5 M( h
Proof:
1 c' }6 n* J1 b8 H! ?! C6 ILet n >1 be an integer ) _8 Z3 k4 F7 k
Basis: (n=2)7 X) V' \2 x( G" ]& {& u/ o8 k( G$ w
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
, |3 _! w6 X3 S7 T2 j. ^( m+ W3 g
Induction Hypothesis: Let K >=2 be integers, support that, L. L$ ^* j$ Z& ?
K^3 – K can by divided by 3.) `: Q6 B+ a! v, Q8 b
( X+ F" [8 o8 H) E& ?# W) cNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3" k# L# |5 @: F9 W }* L! W1 b: {
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem1 r0 S# C/ g# ]- E( r& o
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)$ Z6 q O L; [6 r; C
= K^3 + 3K^2 + 2K
+ ?2 r2 b) _6 y1 {3 I- Q. _- r = ( K^3 – K) + ( 3K^2 + 3K) c* K; @0 ^) E" }4 U
= ( K^3 – K) + 3 ( K^2 + K)
4 i0 }! s& ~3 R% u$ _4 Bby 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
1 R+ B" b" L+ J9 `. O5 n; V$ ?So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)- [0 P; p5 P. S c0 T9 ?' }( m5 |
= 3X + 3 ( K^2 + K). c$ p* B0 U H
= 3(X+ K^2 + K) which can be divided by 3) n# K( h. }2 k- n
$ T! k( b1 X/ T0 B/ \Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
. Z6 W* b1 y) n' U1 ~+ Y/ k7 g: ?6 ^0 l4 K* v
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|