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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)1 q$ E0 N' D0 `. K8 t8 ~
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Proof:
- O( t7 c2 N# uLet n >1 be an integer 7 q8 E- G/ Y! r8 G9 K
Basis: (n=2)& \$ S1 z, N8 h
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# N( s7 ]+ W# o' {! E2 R2 Q
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Induction Hypothesis: Let K >=2 be integers, support that# I6 R+ `1 U |0 \; `0 k0 V5 D2 n
K^3 – K can by divided by 3.+ |0 e5 [* |7 I, z0 N9 W$ f9 i5 i
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 37 C( ?, Y2 F7 {" i4 G- l6 P" n
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem0 P. _* o& ^* ]3 W0 r5 ^# |5 w j
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
9 W0 D9 E; ~0 `2 Z9 j- a& ^& \; o = K^3 + 3K^2 + 2K
* f2 X7 B) o/ t' g = ( K^3 – K) + ( 3K^2 + 3K)( S- I0 l* L' l
= ( K^3 – K) + 3 ( K^2 + K)
0 s5 w9 `: Z/ l3 Iby 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; L4 b$ c3 u& k" HSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)" r( U( ]" [/ `
= 3X + 3 ( K^2 + K)" z# {' U1 [9 V0 S6 E
= 3(X+ K^2 + K) which can be divided by 3
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! v5 O5 g, S4 }8 y1 nConclusion: 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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