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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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Proof: " m: V5 G g4 [' O- `# T
Let n >1 be an integer * N+ v p( M$ s" J" R
Basis: (n=2)/ K$ E: w8 N# \! E# c! M
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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$ h. `, G% q0 P. bInduction Hypothesis: Let K >=2 be integers, support that; s- J7 e1 r1 a1 P3 M
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: I3 W. F% m/ ?& a( `8 a
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem' c5 { K$ J* x+ c# R$ @
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)7 U) Z" C9 J6 N O
= K^3 + 3K^2 + 2K& t/ e- X1 F3 u* J; W
= ( K^3 – K) + ( 3K^2 + 3K)& e N( x2 { ~+ G L ?4 e$ r
= ( K^3 – K) + 3 ( K^2 + K)
: g5 w( s( `& y7 @7 u" qby 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>02 A I: K+ T. ]! Q, {
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" [0 Z/ X2 a# c7 e) C9 z = 3X + 3 ( K^2 + K)1 }! E) j9 S6 X; f
= 3(X+ K^2 + K) which can be divided by 31 @1 R7 ~1 r$ Q% S6 t" }5 H
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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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