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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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% C3 g5 u: \9 E/ D* R/ nProof: + o9 J: O- h) V2 V
Let n >1 be an integer
5 T# N) q8 y# o9 aBasis: (n=2)
" |' Y) c! R+ @0 t7 | 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
- O# ^0 S& M' o; }. l K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3# `0 z! s3 B5 k& ^4 s, B
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
: f l5 z3 q- @. J; l& d! @Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
1 c$ f/ |* K7 A2 Q1 a/ j I! c = K^3 + 3K^2 + 2K
# t1 Z0 n* Q c6 }9 O* W = ( K^3 – K) + ( 3K^2 + 3K)% Q# S- ?$ H* o" H& j6 k: z. u
= ( K^3 – K) + 3 ( K^2 + K)
7 H6 |. w4 H* r8 U; ^0 cby 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
2 ]6 w- K5 E$ a: o- A' k- K) ~So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K), a, x: _& h1 _) O) `' Y0 M. V
= 3X + 3 ( K^2 + K), V. ]$ m/ s0 u1 A. T8 y
= 3(X+ K^2 + K) which can be divided by 30 K1 |4 @# J. j( c1 D
9 ?9 U% B, G$ H- U! U% AConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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