 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)) p! _) B" a4 x8 K0 x3 d3 C
5 |; m7 ~9 G3 ]
Proof: : I2 y1 A+ F/ Y( V
Let n >1 be an integer
4 Z8 q; r" o3 Y6 j+ HBasis: (n=2)
! D! ?4 h0 A$ g6 h 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
! \/ X* Y( U, n9 D1 I9 g+ l" w
; A! X. z+ f$ HInduction Hypothesis: Let K >=2 be integers, support that
, X: ~- F. e* k7 e7 A K^3 – K can by divided by 3.$ g/ y# l7 _8 A) s
) e* m+ k, _# F- L' A- ^4 v
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
& z7 a5 O1 @* hsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
$ M" I& p7 ^" R3 BThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. j0 m" w) B+ k7 S = K^3 + 3K^2 + 2K
. ?1 T& B4 E# U6 e5 y = ( K^3 – K) + ( 3K^2 + 3K)
# J5 Y% _' ~5 e, V. M/ x# G = ( K^3 – K) + 3 ( K^2 + K), {/ B- H: `+ u6 {9 h
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
- d( Y# a- c- Q# w1 ^( N0 [4 X5 uSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)9 b) U; l3 M U
= 3X + 3 ( K^2 + K)% x& c" i @% B! s$ o* \, M
= 3(X+ K^2 + K) which can be divided by 3
1 x u* I! {3 c2 ]% d
* ?# T7 ~# y; Z; d. N; w% j' C6 SConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.% Q- X0 ]8 ^( Q# J
B+ u' _( ^7 }+ Y# v4 l[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|