 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 U6 P1 ~/ F6 L5 `* U$ S
! c& K8 E0 g$ ~2 _9 S# X* u) \Proof: 9 C: |* w, t M2 Z
Let n >1 be an integer
) Q: G7 q' e4 s: y) B4 ~Basis: (n=2)' R7 L, v1 J! x0 C8 \+ d
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3/ l; u* `" ^! O
0 e, p/ U% I. k, j, X9 a
Induction Hypothesis: Let K >=2 be integers, support that
# p2 L0 x1 ^! c8 q; G K^3 – K can by divided by 3.2 A4 e; c& Y+ X; r ?2 L
1 n( o- M* S" f% _0 v
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3+ B* t- m9 z% d Y" r ^
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
~/ P, |8 C/ \& S b; nThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
) ]6 y" v) `+ p5 P1 z7 P = K^3 + 3K^2 + 2K( d) n) N8 b$ P& L$ F
= ( K^3 – K) + ( 3K^2 + 3K)
; V3 j3 t7 z( n = ( K^3 – K) + 3 ( K^2 + K)! F' y8 ^7 A# Q; i( S/ Y$ E
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
, Z6 t, }4 e& X% BSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ @4 `& R) e* O = 3X + 3 ( K^2 + K)0 i1 `4 T U7 I9 c+ k( {
= 3(X+ K^2 + K) which can be divided by 3
: H, e+ ?8 o8 k( W5 g2 x+ O
$ {, y3 i) M9 O/ t! [0 eConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
0 Q4 U2 n0 ]4 X$ k! K1 f4 m* u, _& P
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|