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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)9 o) { D8 \0 D
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Proof:
5 d- {5 R2 U1 M$ R. \Let n >1 be an integer
' ?3 M# p6 G1 D) K' TBasis: (n=2)
& m- h8 l C/ k2 d7 m* h 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3, b6 k B4 C4 q$ h$ S% @, S; w
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Induction Hypothesis: Let K >=2 be integers, support that' y" l3 C3 ` k& U
K^3 – K can by divided by 3.
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$ F, Z: Q% l# {Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 31 @: _- T" S) O _7 e
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem; L$ k8 z7 m4 N" l8 w
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)3 f t* z4 O5 f7 k) j% W# l" l. h7 x
= K^3 + 3K^2 + 2K3 w. P2 B8 s7 d" ?/ Y' B9 m
= ( K^3 – K) + ( 3K^2 + 3K), J' a! v8 Q" r) {
= ( K^3 – K) + 3 ( K^2 + K)$ K; h' S$ x: j7 M0 v r# X9 H
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
( ^- Q/ t( O$ Z$ J' }7 J5 l$ h zSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
7 L0 F5 d7 c, C( z- Q, _ = 3X + 3 ( K^2 + K)/ v& u) F0 l+ J# I' u
= 3(X+ K^2 + K) which can be divided by 3
u6 n6 @, L9 E |! s. L4 P/ U4 ?7 W' s+ K( p+ q I
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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4 |1 G8 w6 D- V/ U[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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