 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)
1 {3 ?9 {/ E0 S" U5 z
, N& v" _1 {0 x, [6 ?Proof:
0 ~# H E+ U {Let n >1 be an integer
" o; c2 P4 S$ h- v G6 F: w; w7 KBasis: (n=2)
; {# ^# ?% v* J& N# J+ d 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 39 E+ j ^+ }5 ^- t# f
U; N: K2 |$ j4 n8 J
Induction Hypothesis: Let K >=2 be integers, support that C& b* f0 E- t. i/ ^2 t2 V2 e
K^3 – K can by divided by 3.
; Q: C: J; u$ \: {" E
2 w' q- D" W* @* F: K" F b" kNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
% @ g- p* D/ Z# X9 f# A5 W- J0 vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, }( b- z1 x* B9 O9 u6 t1 P
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
$ n& h+ z* A3 B( Y$ u g8 q = K^3 + 3K^2 + 2K& S1 @6 T% P* z6 Z. k+ j1 v; \" k
= ( K^3 – K) + ( 3K^2 + 3K)
5 E. f) ?% O i2 ^; L = ( K^3 – K) + 3 ( K^2 + K)' e# P- a1 [2 `! u
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: R: o4 d2 S! E& u
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ c2 N" z6 s! R = 3X + 3 ( K^2 + K)9 R$ O4 Y" f: L/ Q# ~; o6 M6 g
= 3(X+ K^2 + K) which can be divided by 3
$ v/ V0 Q1 @ o3 {
+ i" K w0 Y# E9 G' N) n! OConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.3 `& @" I: z, H/ C
# Z# B; X0 T. _' G$ K7 R
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|