 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)
- ^' r' O3 I9 k9 d) a
' J9 B0 x: U1 o1 AProof:
, h0 A5 K; ? A1 s& m& QLet n >1 be an integer
1 v+ @, b2 t5 e [. ?; V: QBasis: (n=2)% S7 V. ?& H- M2 B' y) Q2 z
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
& V5 B! t- U2 G+ Y* z2 w; d2 P& D/ u# u. {
Induction Hypothesis: Let K >=2 be integers, support that
- t: X3 `1 `' k1 { K^3 – K can by divided by 3.
1 o, f1 N+ y' C# e7 g' a) ^! F/ B% G6 k: w% I
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 35 ]4 |! h3 V6 h( C% i0 I
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
+ e& }6 ~3 q# Y" o8 F4 GThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1); v6 g& j- J! Q+ J9 G' h3 m9 H
= K^3 + 3K^2 + 2K0 K1 ]( R, b0 \ P( t8 v
= ( K^3 – K) + ( 3K^2 + 3K)
% I4 M! P: {- E5 |5 l = ( K^3 – K) + 3 ( K^2 + K)) _( `% Q7 m# c7 Z
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>06 }9 T9 j9 Y3 ]& P5 w
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)& j5 ~/ Y9 G3 @# _
= 3X + 3 ( K^2 + K)
0 C8 x0 V! L; O: ~% f8 O, ^& C = 3(X+ K^2 + K) which can be divided by 3
3 ^$ K: }2 `8 ^% q5 n
& f1 _! m4 S0 ?% J# kConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
7 w7 s+ f. M6 }) X" l" t! d" K4 }* K5 S* Z' H' X
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|