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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)
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& w* z2 v: [6 AProof:
- g) N/ V, j+ |% k. q2 Q; oLet n >1 be an integer " r; y- {6 a$ b1 | D2 ^( s
Basis: (n=2)
4 z% T- w& N5 r* J3 F7 W 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# Y) c0 m! |/ M# D
3 q7 ]5 Y( T) ]- P+ X
Induction Hypothesis: Let K >=2 be integers, support that5 {- ?+ ^* W7 t* C$ g4 n; w
K^3 – K can by divided by 3.' s! _" c2 ?0 g: s# A
" _) K% W3 c! f$ w1 W8 q$ @- RNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, n4 u0 W% {) z r8 Rsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
# J0 ^1 |, Z' _$ }Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
# Z% t6 P8 _5 i# l. c0 C- E$ i = K^3 + 3K^2 + 2K
! _$ J" S% |- o7 f+ ~" y* U6 n& V = ( K^3 – K) + ( 3K^2 + 3K)1 p1 {. E) t( _6 A& P3 X
= ( K^3 – K) + 3 ( K^2 + K)
7 z1 v- A) e) K. c* zby 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/ P+ G! {; |5 q5 _4 \
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K): H$ l/ i+ x, I0 j. @" @/ s
= 3X + 3 ( K^2 + K)
3 Y9 w$ B9 d& T" x7 X7 K; v = 3(X+ K^2 + K) which can be divided by 3
+ G; ]5 {6 Q! y$ K* J; F4 S
3 j7 X" d7 ?$ h5 ?4 [+ ]Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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* M) J: }. K( w: R# N[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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