 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)0 y) c6 a$ b$ i; l" H4 ]
5 _$ Y1 ]6 S9 @Proof:
7 }0 n' s7 _+ Z MLet n >1 be an integer R, j) x+ H, A
Basis: (n=2)
& z7 D6 R6 o! M1 p! C 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
' r# W- D; n$ D; a4 ?
- S( g: j9 J& ~8 E. mInduction Hypothesis: Let K >=2 be integers, support that
5 M# a9 }1 w: d7 I3 S3 ` K^3 – K can by divided by 3.
/ o+ t; x" b, \& N: s2 i, u' C8 ^& f. o; a H# A
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
9 k5 X. @# R. F7 n3 G+ ^7 ?since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
* f9 R5 E# T3 @; I8 v" ^+ iThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
7 }' A' S3 o1 T7 U; r } = K^3 + 3K^2 + 2K0 H) A7 L% K& Q8 F
= ( K^3 – K) + ( 3K^2 + 3K)
! M c( K& W6 {, @/ ? = ( K^3 – K) + 3 ( K^2 + K)
) n* ^! A1 ]' r7 sby 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& W! t5 a5 g6 E" x; A0 l) p
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) N/ @. @! W7 |+ \' s% Y = 3X + 3 ( K^2 + K)
; v8 w; D" s! P) ] = 3(X+ K^2 + K) which can be divided by 32 n) I& B6 \/ e) p, j$ i, I
7 ^6 W3 j7 w! w6 @% A5 T7 J" }
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.5 Q' F( b% }) d7 e4 t- S, T. a* o1 ~
2 G7 Z) ]$ a) x% c7 n
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|