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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:
) R% E$ z2 Z# N2 d" {( w! v) ^Let n >1 be an integer + Y6 q% X9 a* B
Basis: (n=2)
; C/ Q) w/ e+ y 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
' s6 ^( X) F7 X( q# d6 V# O! J" ^ K^3 – K can by divided by 3.
5 J# x3 ]9 i \* e K" l* G$ a* L3 T# s) O4 p5 C) |; F8 ]" r1 J% S
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
5 Y8 @! F( c6 K8 Jsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem7 ]: G9 Z2 [4 |4 m
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)2 U: d( k8 m; a4 F
= K^3 + 3K^2 + 2K$ U9 `( c6 B9 u1 \
= ( K^3 – K) + ( 3K^2 + 3K)
# Y; U! Z3 X3 Y7 ?* ?; U( ?7 Z = ( K^3 – K) + 3 ( K^2 + K)' k2 }9 l) i' ~- ?6 R" 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% T- O$ h' m) |& i" N
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)! `$ j5 e* W1 Z& b; ?' |
= 3X + 3 ( K^2 + K)
8 N; v1 ]. S, f* Z3 l2 k! o; O = 3(X+ K^2 + K) which can be divided by 3' f! Y! S8 j2 G7 _) m
- U# X/ [" b+ N6 y
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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