 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)
% W' y# S( C$ L1 M) E' S
4 S. n1 D4 @, F# U/ MProof:
) }! o( b8 }0 n. i4 L8 d# K/ sLet n >1 be an integer
3 [# {7 S- I1 P$ F8 B. fBasis: (n=2)/ Z3 E: ]$ v' c, \4 a0 o
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 37 m5 O3 i x3 t
! V2 F0 E9 a, ^ s, O9 Z: B; b
Induction Hypothesis: Let K >=2 be integers, support that
" c f9 O9 m! O6 j; M. K K^3 – K can by divided by 3.
- T' J: S( G a, x4 X
' e( e G: V, _/ [# M* h) BNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
& f* R- ^( l Hsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem) N8 @' o8 ?2 m' E9 ?# F. c
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)0 F3 K% p% F3 G, S, g& O. N. L
= K^3 + 3K^2 + 2K! j( B* @$ d! o g/ X
= ( K^3 – K) + ( 3K^2 + 3K)4 |4 [% f( `% Q) i7 z, O
= ( K^3 – K) + 3 ( K^2 + K)+ j# J3 \0 z/ T: \. K! E9 T
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>04 X/ t& _. o3 f4 M
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)8 G8 h, _& b8 y( y. c
= 3X + 3 ( K^2 + K)
9 i6 j5 V! j/ \1 C/ V" n = 3(X+ K^2 + K) which can be divided by 3/ z7 K' \+ j" @
3 j1 g/ r- c; {Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.9 Y8 F* N5 K9 f
* q0 I- N( n( C Q[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|