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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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. H8 R6 k) z- Q p% cProof: 4 w3 b6 n9 T# u4 s! D
Let n >1 be an integer 9 \0 v3 D" c: E( p8 T
Basis: (n=2)
) R3 G9 ^$ n. M6 X 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 37 R" t: f; \( @6 W# p) h, S
1 ^" ?, D# F8 Z$ X/ y, |Induction Hypothesis: Let K >=2 be integers, support that; k9 c8 u) T- A; G
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# A/ Y( o- F9 a) O
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
6 ?: l V* U: U4 @3 Z; O7 _& TThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
$ v; j: B4 Z8 o% U- i = K^3 + 3K^2 + 2K
( D8 }; N' F: i& T; x- D = ( K^3 – K) + ( 3K^2 + 3K)" @6 ]/ {$ c4 V, Z N8 r( s& \+ O
= ( K^3 – K) + 3 ( K^2 + K), n8 e& U8 c3 d0 b7 ~; L6 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) q" D7 Y" t# k, L: u5 B6 p
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
8 f8 o( j' w" F0 \# _4 o! }, G5 k = 3X + 3 ( K^2 + K)7 \( J$ Q- K, \2 R' ]
= 3(X+ K^2 + K) which can be divided by 3( v1 ]$ N$ u- d. k
+ H( C; c6 N, n- w, _! a7 k: A0 LConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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% H3 X* B7 Z0 I9 ?9 ^1 S' S' }[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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