 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): M! O5 l1 w6 v3 T" m9 b
$ R. M/ }8 v# T3 y& F/ G: i3 `
Proof: 7 e5 ^' y% u5 f. F* J' }1 `, T
Let n >1 be an integer
) j: m" y4 |2 H1 G! c) dBasis: (n=2)
% z; F! [; l B- ]6 _, G 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
4 ?! W, Q+ |& v; m' D+ M, Q% S% ~, [: W% F; U- j, T" `' S! c
Induction Hypothesis: Let K >=2 be integers, support that
% H$ S* S7 {' O" z K^3 – K can by divided by 3.
5 s$ H5 \" y8 h% z: [" s" p! d4 b. r# c; t# l# F5 [" A. y# S/ R
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3/ C T& a( {7 c5 T% K
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
+ B) `$ {3 y* R3 J1 e, q! KThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)+ t+ U; K" g% k6 ^' w* T5 B4 t/ }
= K^3 + 3K^2 + 2K% m+ }0 J/ g2 B5 }
= ( K^3 – K) + ( 3K^2 + 3K)
% V/ t5 q: t) C& H' ?, {5 | = ( K^3 – K) + 3 ( K^2 + K)$ N' T' W( R* u# N; w1 H: `
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
6 T I9 @5 X5 C% o4 y* ?So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K). ~6 c5 x( R9 D3 h) }) e! n" ?
= 3X + 3 ( K^2 + K)
4 f3 i/ z1 ?6 G2 Z7 ~0 K = 3(X+ K^2 + K) which can be divided by 3
9 p& \8 c+ [+ G9 ~/ ]# D. W8 T; C3 K" |9 ?. l8 b
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
" G G3 B5 d* a. h& V/ u0 w0 E! P9 G1 k9 g Y8 f
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|