 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), g2 G' E( u7 F) X* O6 v! _
* @! w" Z5 f! K2 Q8 T7 d
Proof: ' V# J* a6 t7 R6 W: f
Let n >1 be an integer
# V2 @ t1 D4 d5 vBasis: (n=2)' Q6 C, h- p5 J/ f8 O! Z
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 34 y6 ^5 s6 I, z
1 M( W# v& i. o `: t3 p4 c
Induction Hypothesis: Let K >=2 be integers, support that
3 \3 A+ d- D- w6 Z6 ] K^3 – K can by divided by 3." o1 q$ ^; U' I" ?5 Z5 m
2 g& j1 U6 g F5 N! k9 kNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 32 G( g9 G( q4 v
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
" ~' d' V4 Q: |" UThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1), O6 p" M4 ?1 ?" Z9 ?
= K^3 + 3K^2 + 2K' f$ w V3 o; B2 i! a
= ( K^3 – K) + ( 3K^2 + 3K)2 f' j& B0 @9 O8 R! `" [
= ( K^3 – K) + 3 ( K^2 + K)
4 x6 x* W* T' P. l3 G3 F: Mby 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
, S d- _: Y$ F$ J: T' x/ p" a# USo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)0 J% Y% n. t4 B' r
= 3X + 3 ( K^2 + K)+ k2 e7 J1 H; }
= 3(X+ K^2 + K) which can be divided by 30 T6 q' |- k; u! t: d5 n$ q
/ ~9 O* \: y5 w$ I: R% |7 b* ]Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.+ }7 l. z9 C7 i H# _9 ^' G
9 {" D0 A3 ?$ j$ a/ v O[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|