 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)8 y( h; b) F9 T6 Y: @4 v
! e3 O8 U2 K4 s& M
Proof: % `% a, Z: b" Z; P
Let n >1 be an integer + }% i6 I# B5 W, d
Basis: (n=2)8 v2 C5 |* N6 t
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
4 S3 x1 m9 s' c& }5 I: J9 b- _: E4 D* g3 \
Induction Hypothesis: Let K >=2 be integers, support that7 }( V8 a. y7 _/ s
K^3 – K can by divided by 3.
' a8 N9 I) A; C
4 K+ D, ~8 B6 K0 E# H7 ^Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, W; q& @9 ]. @7 c) k& ssince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem; y4 o. a8 @9 \4 h- \' N& i
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)% m3 `# l9 @+ M i7 v$ A
= K^3 + 3K^2 + 2K5 ^. H1 O- f! z w; l2 X! j+ f% F
= ( K^3 – K) + ( 3K^2 + 3K), \ W% v3 d1 ?( g
= ( K^3 – K) + 3 ( K^2 + K)- ` p; O* ]9 \- b( f q
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+ y# [: w+ C* L9 g& U
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
7 P( s/ h2 W2 b; r' M* o = 3X + 3 ( K^2 + K)% A8 M0 a2 x: r8 B5 G- E
= 3(X+ K^2 + K) which can be divided by 3: m) i; Y( {( `8 \. c1 f: F4 K
3 |& o- r% W1 d9 WConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.7 J6 z1 D! D6 q$ i9 j0 d
3 M$ M2 R* x* d# W. X) C[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|