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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:
8 z! V- v M+ HLet n >1 be an integer ( O3 I5 ^ s' z7 ?* X+ c# p
Basis: (n=2)
2 R! u& S# O$ U0 M 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3. E, f5 i A {& \1 d) a
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Induction Hypothesis: Let K >=2 be integers, support that
. B' a+ z" |' n/ e K^3 – K can by divided by 3.
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4 s! \9 `$ c. f* N( yNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
8 ~% Y6 L, Q# y- `; w! f. Vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
6 T: s" ]8 X2 O2 H9 b4 qThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
1 r$ U) C6 Y/ B8 G' h = K^3 + 3K^2 + 2K
. |9 L$ `4 }2 [9 g5 u+ v8 i3 p = ( K^3 – K) + ( 3K^2 + 3K)
$ Y' ~$ {- q! J `8 n$ T4 O4 Y = ( K^3 – K) + 3 ( K^2 + K)4 K* I2 T7 ~: H) p/ K
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( [. q1 K' y6 G* w5 }7 E* R
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 p8 Z5 F& W6 P" q: ? = 3X + 3 ( K^2 + K)
8 ^# e; e" u" C5 q" l = 3(X+ K^2 + K) which can be divided by 31 V4 q1 A- K% H: Z
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1./ K8 i# y" W& f+ t; Z6 O, o
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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