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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:
) y+ W$ n9 R! e5 SLet n >1 be an integer 9 i+ f/ x' L8 w7 E/ U9 t; f6 N
Basis: (n=2)" O& d% n5 l, c, }
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3" I8 R- {: S* E4 W9 F
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Induction Hypothesis: Let K >=2 be integers, support that- N2 i- N. w9 M. Z/ U* z. L, R9 e% e
K^3 – K can by divided by 3.; g a; a6 V! _5 C1 {
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3" u; p9 U8 B& N, {* i8 p
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem4 o7 V5 o& z- a$ T
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)" j- v$ ?8 ]; M) \4 ?$ b# Z& ?
= K^3 + 3K^2 + 2K s+ e8 I% e" J6 O
= ( K^3 – K) + ( 3K^2 + 3K)
9 ?% r2 [: |6 E6 H# @ = ( K^3 – K) + 3 ( K^2 + K)
+ \* K; p2 c: u; {5 uby 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
+ M. ~: g! ~& w2 n3 `So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)( N3 o& v3 ]; c% f z) V
= 3X + 3 ( K^2 + K); S v. X7 Y" _4 v( K
= 3(X+ K^2 + K) which can be divided by 3$ m X2 j5 D7 s8 D. H
! K; v0 {9 x. K+ m8 y, w7 \
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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3 X! L) ]+ w1 U3 W7 [& N[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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