 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)5 T" D+ j; ]$ U+ C
; J4 C3 _) t! _2 _, C
Proof:
# U* x4 x& L' ?8 y' aLet n >1 be an integer 1 [6 W/ s$ ?3 Q
Basis: (n=2)
# J5 C3 G D) Q9 x4 k6 R, A 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
3 V5 Z6 H5 p9 Q( U: I+ _( _2 m& l5 F0 ^2 D# e" J V
Induction Hypothesis: Let K >=2 be integers, support that* y/ c( K2 Q4 O: g, |- f
K^3 – K can by divided by 3.! S M" Z6 N; t
$ ~3 _2 p: I, E$ t
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ t) y$ N1 S8 J/ L: Y# p* Y2 Hsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
9 f" G; ]) O, x: oThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)4 F3 R6 m5 G" m% p( a2 t
= K^3 + 3K^2 + 2K7 s4 g/ O4 _' L, o; p
= ( K^3 – K) + ( 3K^2 + 3K)
1 X* f+ D. J8 V" c = ( K^3 – K) + 3 ( K^2 + K)1 i3 o# P7 R- v/ p6 I
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>0' H( J( l ~4 d& e1 p5 C
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)5 W- l0 i x( o
= 3X + 3 ( K^2 + K)
$ a: @" n. E1 a0 @ = 3(X+ K^2 + K) which can be divided by 3
. I5 C( G [! l7 a! b; T3 ^( \! b i$ D M, @
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
8 H* o5 o" \* n( W( ]- A2 w0 O6 |* l9 w6 y1 L! ?. s. E
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|