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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)
7 n g9 j c0 b; A' {4 r4 b& d
' i1 M# i" E s: l$ ]4 yProof:
7 l! T2 l& n; I& M ?, j; JLet n >1 be an integer
# t5 p W J) N s3 lBasis: (n=2)
8 W2 _ r8 z9 h$ K$ { 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
L) W! k! E# g# w2 N
( [$ f: f+ x9 g6 h7 Y: TInduction Hypothesis: Let K >=2 be integers, support that
4 J1 O( n* P3 f2 v K^3 – K can by divided by 3.
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|" k2 b/ W6 L' ?; P% o( _- gNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3( m& G# u+ l% r0 K# T9 I7 L
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, m. H5 z) @: ?
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
6 Z+ w, r- E3 f, k) B9 }& n = K^3 + 3K^2 + 2K6 L, S _, f5 L; J3 g' n" ~
= ( K^3 – K) + ( 3K^2 + 3K). V! ~/ N2 h* n9 J/ p# R' h
= ( K^3 – K) + 3 ( K^2 + K)" w5 D3 Y- R- A) ^) S3 B0 [
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>03 @ m0 Q3 U4 C% ^" W
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)1 z, u& L( G% s' i# U$ w
= 3X + 3 ( K^2 + K)3 C3 `& i" y( r) Q9 z p
= 3(X+ K^2 + K) which can be divided by 3; V) N) [. f0 L4 D; Z4 u
) n* W5 Y/ d& W4 a% o7 j G
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; a4 ]8 `% w- Q$ X
2 X' q. ]" c8 E4 ~- O[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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