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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), P2 S7 _- J1 L" R
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Proof: . v: }6 G. B! s) J% o
Let n >1 be an integer 8 P0 o" Q6 B! K4 [8 m7 v
Basis: (n=2)
, {; ?+ h2 m2 ?$ o 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) H0 m I/ G. I
K^3 – K can by divided by 3.
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0 K! ]- l0 S. A) W" b- y& M6 {Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3" p* D6 e% D; b; K2 ?$ t9 t- F
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem0 c2 H! Z9 r( Z! v/ F. m! z' I
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
1 O$ a, Y- A. H( m' r4 l5 e% ~ = K^3 + 3K^2 + 2K* |/ `! g) v) L2 I: p% k
= ( K^3 – K) + ( 3K^2 + 3K)
: }% a4 U+ ?3 ? = ( K^3 – K) + 3 ( K^2 + K)
9 k* {) ?7 `- |! K' \$ oby 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
8 k2 |( x0 F7 K5 w* aSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)" K, F( e& O/ {: X) c
= 3X + 3 ( K^2 + K)
y" Y. p) h6 F! Q0 \5 e8 S5 ]" e8 H = 3(X+ K^2 + K) which can be divided by 3% g0 f! Q9 n6 t9 X
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.& C1 T+ L' r% P# Q
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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