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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: 5 \8 R% z, F2 `7 ^
Let n >1 be an integer
. G2 T. ~$ x; T9 [* a4 w# ~ e9 q$ KBasis: (n=2)
+ a* {; Y/ m- c3 o7 E 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 34 D* W- l6 k. U1 V
1 S6 y/ Q, x, Q+ z8 zInduction Hypothesis: Let K >=2 be integers, support that4 ~* l; V5 H7 J
K^3 – K can by divided by 3.
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* P( o7 q: m: t% L9 I* `Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
3 N% X0 x1 a$ e& R9 p9 Vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem8 I' u7 F" g& }2 P4 f
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
1 {$ x. d7 z a = K^3 + 3K^2 + 2K; f7 S4 ]6 O& J- Z6 o$ C7 w% @/ J
= ( K^3 – K) + ( 3K^2 + 3K)5 N* j {7 a: f6 w1 d
= ( K^3 – K) + 3 ( K^2 + K)5 R/ k& B1 ^) a6 R( x2 I
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& R! Z& O, a8 Y$ ?9 U/ ~
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ w% k2 p5 n! Q1 B
= 3X + 3 ( K^2 + K)
( X& f P4 Z" @ = 3(X+ K^2 + K) which can be divided by 3% y* {- Y4 U0 C6 m/ ^( ^
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.+ w3 _+ `* E7 h$ \
, U3 S% V. d8 M# ]# r[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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