 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)
. D/ V3 J; N; A8 z% s8 \ R- ` Z2 y: q9 O
Proof:
( j# A% E2 J. O& M; Q1 YLet n >1 be an integer
& {% f1 X- l. v wBasis: (n=2): J% V9 j! e, F5 q. k' D
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
/ q7 ] z+ w+ [3 B4 x4 q; v
6 x; o6 H- @% l" i& B+ mInduction Hypothesis: Let K >=2 be integers, support that- ~! O7 {6 F. @: R# W+ @
K^3 – K can by divided by 3.
0 E. E: w( z/ _4 S c5 b2 T& Z0 i: ~3 @
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3! u: p2 I. ]) w& W0 f
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem8 l# C, [8 N9 o% g1 Z2 J
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
* c! g" L5 w9 ]0 ]+ Z( C5 S$ Y+ Y& u = K^3 + 3K^2 + 2K
) Y& e |; V& |$ ~. `+ c2 A = ( K^3 – K) + ( 3K^2 + 3K)& K, ~2 n/ K9 [$ a+ T% V( M/ \
= ( K^3 – K) + 3 ( K^2 + K)- N/ i! N7 Y% D% @7 t s! @
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>07 ^; ]& R7 s( O+ D* ~$ D6 \9 T
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ ?, V2 G, m, y, u6 m9 h = 3X + 3 ( K^2 + K)7 ^$ R, u: N$ r2 m* @ M2 O5 s
= 3(X+ K^2 + K) which can be divided by 3
/ \4 X$ t+ }2 x; h9 T4 `9 r B S, x, V- Y6 |
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
. p; J* T' l- d; w1 W5 y* L$ _! p2 ], K* l5 B( H" K- E
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|