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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:
* [1 u3 x; B& A, oLet n >1 be an integer / N [( x9 _1 i! d
Basis: (n=2)
* t$ i) `& ?; `5 @1 F* N 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 36 N* r0 ~) F' h
- {7 Z' q( A, e. j( E& k+ B4 VInduction Hypothesis: Let K >=2 be integers, support that! j2 h: @/ y' e2 I) }
K^3 – K can by divided by 3.
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7 z. h, r% g( L5 L; x; @Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ {9 V. G# O, ~1 j4 ]7 Usince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
. o& ~: {$ x/ V( s0 XThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
: {4 T. E. ]2 ~- y- S = K^3 + 3K^2 + 2K3 ?* ]$ v0 u+ p0 u. S
= ( K^3 – K) + ( 3K^2 + 3K)
) b- S; M- u6 g/ i( @4 ? = ( K^3 – K) + 3 ( K^2 + K)
7 @% [* C5 }# n' Bby 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& K/ i8 j! j# h1 A3 q+ ?& G
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
5 {# N# h. K4 x: s! P; | = 3X + 3 ( K^2 + K)
( }" H) h ]" D8 K" c = 3(X+ K^2 + K) which can be divided by 3+ ?6 N/ M2 @, l U3 D% F1 a
$ K" Q1 R, w0 V0 y6 A _- m$ t% A1 P7 @
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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