 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)
9 M& r) r9 _; K2 x
h3 ?- A2 a1 U: _) kProof: - R" @* S# I2 a+ s Y
Let n >1 be an integer & j- _$ q+ n( B% u; k3 B" X
Basis: (n=2)
5 D1 v! l% b6 y6 V( j 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3& C- ~9 w- { o) v& `8 c# m9 i
" v# B8 X) ^1 j) _Induction Hypothesis: Let K >=2 be integers, support that
# N8 l) z3 S5 k# _ K^3 – K can by divided by 3.1 D; [( f( u+ ~4 C3 a
8 J& E# f8 r7 x, b( CNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
$ Y/ i, ^# x0 a) hsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 U4 T1 o0 X1 U# gThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
* U. Z/ t/ J1 A' x e/ o) e = K^3 + 3K^2 + 2K
$ a1 P1 a$ b/ h8 Z$ \$ T, W = ( K^3 – K) + ( 3K^2 + 3K)
% ^' g2 ~: K" y# T = ( K^3 – K) + 3 ( K^2 + K)
5 Y0 t9 Z- |( h2 X" }: I, 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# D. u2 a) `1 }' y7 v( l+ e4 G
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)( G7 k! b8 s1 [4 P" u4 d9 G& f
= 3X + 3 ( K^2 + K)/ k3 E. X8 f8 Z2 u. q
= 3(X+ K^2 + K) which can be divided by 3
0 i, ?8 a4 q/ E* G" L6 o1 U( b R# Q% b; _: y9 W6 Q8 S/ h# J; D
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.% |# Z- X: K) S0 Q, k4 R' g
- }; j" U' S5 E$ D9 I, p[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|