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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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6 X1 r5 d% a8 I- ^& h8 R S) NProof: 2 v4 _/ B# Z% J# j& {
Let n >1 be an integer
# c* S& J. b9 U# \( T. D, yBasis: (n=2)" c4 j2 v* ]6 K" W
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that/ a9 I5 J; G# O6 a/ \$ @5 I
K^3 – K can by divided by 3.4 M9 u1 Z; C1 b- D4 j
: w4 D% t4 s8 ]! N. B" k
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 32 G$ D! V* p0 d5 s. \0 }5 i- k0 m
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
; l" {1 x& B1 Y- T" H2 qThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1); `8 e$ p" l3 t5 |( k; L) R
= K^3 + 3K^2 + 2K- m$ S) ^: U K. J3 Q% C
= ( K^3 – K) + ( 3K^2 + 3K)1 F* J' _1 ~! K- q
= ( K^3 – K) + 3 ( K^2 + K)7 Q9 |1 Z+ Z+ X9 A1 y2 I+ z
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
9 m' L' i8 h6 \So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)' g# }5 ~, K4 M+ Y
= 3X + 3 ( K^2 + K)
& d- ]' c" p X' l# }8 H) j6 | = 3(X+ K^2 + K) which can be divided by 37 K G/ t. g" u5 w! [# ~
! N' P5 d( w7 a. |+ JConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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! j8 T- c0 y j2 r* ^% |[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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