 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)
' n% F4 e, V, q
& m6 {) Y7 ~+ O0 L8 i! X7 oProof:
) a8 S, C. ?4 J& E. e0 Y+ y+ GLet n >1 be an integer 6 h P4 ~1 F( y
Basis: (n=2)! w2 V2 d# P7 c Q9 i
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
+ C% V1 s! [0 V
+ M) R, E$ z) ^Induction Hypothesis: Let K >=2 be integers, support that& ]$ Q* C/ k' E* h( G+ T' i- r
K^3 – K can by divided by 3.
E" M$ N" n) j6 m, o! V$ F6 x$ v: {$ ^) s. i; f# L* }' ]* w
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3- D; E, c0 ^* p. r1 a3 n6 U
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem7 r7 ^5 q8 M" }) ]
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)( k" G- T% J1 W: D, q/ {: z; ?
= K^3 + 3K^2 + 2K
- O5 K- @5 C* G( H9 ~ = ( K^3 – K) + ( 3K^2 + 3K)+ w) g9 G8 M* s- n M# Y
= ( K^3 – K) + 3 ( K^2 + K)
2 C' p) K: \4 G1 d7 O# [7 W9 K0 Sby 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. v% J$ |9 K' D6 ]! P8 [2 D, x2 g3 ~9 f
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
, \8 L% k6 J0 e8 @' R& Q: S8 J = 3X + 3 ( K^2 + K). D" c6 l @0 b7 o% S. g
= 3(X+ K^2 + K) which can be divided by 3
( W# y$ \# `& h2 Q2 E
% p8 ]* B$ X, e0 gConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.1 |6 S# s, `) E9 p
: Y9 Y1 d% Q: k( [[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|
狗仔卡
发表于 2005-2-22 07:58
提升卡
置顶卡
沉默卡
喧嚣卡
变色卡
显身卡
发表于 2005-2-22 09:14
