 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)
6 Y3 W" U5 A* S. x. {) }6 ^, G: j6 t. {
Proof:
( x) o' r k1 b6 |, H4 oLet n >1 be an integer * ?0 n/ W) f- L+ q- G
Basis: (n=2)6 M4 L* Q) Z6 |
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3: W. s8 X" F1 A* S( b
9 z0 k2 M* O1 d& n" _3 o3 T, nInduction Hypothesis: Let K >=2 be integers, support that. k0 r1 E6 X$ q, p7 J) c* L& `
K^3 – K can by divided by 3.9 W! r# I5 y4 z
& _. Z; E- Y/ f+ H$ d( n! ~
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
! w2 a0 R8 V: @1 ]+ U) ~since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
^7 j( k2 p5 C+ [ s6 [Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)- i$ n7 ?& I4 O9 P- t, g
= K^3 + 3K^2 + 2K& {" E7 g! p9 N. y, s+ Q
= ( K^3 – K) + ( 3K^2 + 3K)) K3 J' @1 r. ?& k& P# m
= ( K^3 – K) + 3 ( K^2 + K)6 y: x" S0 [5 a/ x4 y) c( a
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
6 j* _) e5 \- P2 lSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
- f5 d5 Z% u8 K8 b8 d$ i = 3X + 3 ( K^2 + K)9 X1 `. j7 U {/ h5 K- D
= 3(X+ K^2 + K) which can be divided by 3
8 v- R! ~' S X( A/ I0 Q8 n$ d/ A' q$ f% g# H' x1 W2 I' z
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.( r# v9 |4 h, Y! Q) G
1 `: ]6 Z7 N; c! \) [[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|