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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)2 s# [) |* i. F+ }0 e" `$ o# k
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Proof:
1 d/ S1 q& T8 S' \: Z! ~Let n >1 be an integer
; k* p: o' C; x7 b* n2 d) cBasis: (n=2)4 h1 E0 h; s9 E2 L
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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( }7 s6 D* X' g6 DInduction Hypothesis: Let K >=2 be integers, support that K+ j: ^% y5 T# o) j7 K
K^3 – K can by divided by 3.& s: `# a+ V2 l% S0 X
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3; Q' i/ \, L8 G1 [6 {, C1 o
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem3 q. e t5 c; ^: o" a% i' b
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)8 y ~( }( n6 z5 \0 {! `
= K^3 + 3K^2 + 2K O. Y2 N9 h3 G7 k: H4 ?
= ( K^3 – K) + ( 3K^2 + 3K)
& o( Q1 W5 ]- p% @2 B# V = ( K^3 – K) + 3 ( K^2 + K)
/ W8 s, ~& u8 [' l3 kby 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( m+ i" o" ^1 |5 \: L% ?. G! T
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K); M* J9 L; V0 _8 W7 z9 {" |
= 3X + 3 ( K^2 + K)) |+ P0 h0 N1 r5 }
= 3(X+ K^2 + K) which can be divided by 3
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2 H* B+ M1 e+ [' eConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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$ x' c& G5 L8 i' x! W6 w9 P1 M[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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