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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)( _4 h1 l7 b6 |# \8 v* h! |; u# i
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Proof: 3 S, l& `* n8 ^( p
Let n >1 be an integer 3 ^/ @# n( b4 n( j( q& }
Basis: (n=2)( ?0 N! l9 s5 o9 ~ H& M
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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1 `' ~4 s/ z5 v5 C4 ?, t. eInduction Hypothesis: Let K >=2 be integers, support that! s* l% L$ G! M' s, l7 M( a$ H
K^3 – K can by divided by 3." }: }* J$ i v
. Z4 o' v% K2 e" qNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3( X0 n- n8 F. g
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem! y) ?' |1 J( Y- K
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)( Z- J+ [2 v8 ^$ v' F
= K^3 + 3K^2 + 2K) Q7 [) p# T9 ?
= ( K^3 – K) + ( 3K^2 + 3K)
( v& i: w& n$ R" a0 @3 E = ( K^3 – K) + 3 ( K^2 + K)1 R c$ L$ Q9 ?, v( T. V, v
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
0 w* O+ q5 B! P s/ fSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)+ i4 l. E: q: h9 {4 I
= 3X + 3 ( K^2 + K)5 t* i C; |+ Y9 ?$ |
= 3(X+ K^2 + K) which can be divided by 3' K" t& A3 A" G, \8 o( N
7 m' j: V9 B y- N P& `Conclusion: 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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