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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/ j$ [# [( g" o+ Z$ t9 z1 p
Let n >1 be an integer
/ B1 V2 @: ]5 T" S& N Z+ yBasis: (n=2)7 H; ^9 [& M$ [1 l8 a1 @
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 32 a! r0 j) a# b8 j+ F
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Induction Hypothesis: Let K >=2 be integers, support that9 w2 K# i9 j% w) m9 Q( Z5 P! w
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
- x4 d3 [8 x+ \2 I0 d* M) Asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
. ]4 L( R. I% h& I( A+ ZThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
; W* t& b4 T/ g$ z1 q* Y = K^3 + 3K^2 + 2K
. `7 R$ M$ w3 D$ W = ( K^3 – K) + ( 3K^2 + 3K)
" D) h: Y7 v6 R% S+ Y = ( K^3 – K) + 3 ( K^2 + K)
: K; V# |7 x# [' vby 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
" O0 A: C; W5 {8 O% S$ eSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
$ I9 v, _' }1 Z! v' u: x8 h = 3X + 3 ( K^2 + K)6 T2 J; D9 r* V% o% [
= 3(X+ K^2 + K) which can be divided by 3' e5 }# Y, J0 m8 H9 c) Y; V2 U
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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D) M* O# F! v[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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