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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)# {- q1 i( c# F
1 R4 [/ u& O" u, n% b9 h6 m2 oProof:
4 N1 H1 O( Y# n& {$ _Let n >1 be an integer 4 H s1 A2 y( D5 C
Basis: (n=2)& R3 a7 z, i/ @2 e+ g
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
8 j, g6 i$ X4 t( q. G( N! N7 \( f( i
Induction Hypothesis: Let K >=2 be integers, support that
! h3 v% o0 n6 ^. k# H2 } K^3 – K can by divided by 3.
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+ |6 K4 _( \, k" h- Y0 aNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 34 S! r: z' z) y& D
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, V. T& p: N/ l$ Y
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)9 f& e: ~ m( y) _ H
= K^3 + 3K^2 + 2K* B2 L6 ]" p* z
= ( K^3 – K) + ( 3K^2 + 3K)4 g i/ K# |+ v4 L0 J9 E1 ]
= ( K^3 – K) + 3 ( K^2 + K)
- @" H4 _& I. h* P: M* Tby 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! ]) C) [6 N& f6 u2 ], g
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) e5 V* c4 l; ]4 z = 3X + 3 ( K^2 + K)
5 {+ p3 a8 \1 E: ~ = 3(X+ K^2 + K) which can be divided by 3
7 [. v+ V! U* [- `8 Y2 ~
* s7 T: X& Y% Z5 N" UConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
1 N2 Z- H% Y; o$ Q& E
% F1 ~0 {7 ]9 m- c0 t1 O[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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