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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)+ l9 p! @& P' I8 l. h
( ~* v1 b& l& W! Q+ ?; H3 FProof:
0 S1 b, b' `9 l6 I/ K) Q! KLet n >1 be an integer
" n- p6 t/ A3 h x4 {' ~, r( C. NBasis: (n=2)
2 Q6 }( r$ A. N0 R5 x$ s' b! ^ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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9 q' T; O0 L, hInduction Hypothesis: Let K >=2 be integers, support that( G* i' S" P% N0 ~4 q
K^3 – K can by divided by 3.1 g$ D! i$ M( G- j
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3: ?2 h5 A6 v) x" G
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem' P& A/ @4 X( \( m2 f5 D3 j
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
' h- V" d, Y2 c- D) s = K^3 + 3K^2 + 2K/ e* r! K3 A% I
= ( K^3 – K) + ( 3K^2 + 3K)5 b3 ~3 G, Q; s5 Q! F/ Q9 ^
= ( K^3 – K) + 3 ( K^2 + K)
. Z' ~. @- R& k D% L+ L5 f7 m* O/ Tby 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
( l B+ q: ]8 O% q" X4 MSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K); e5 {1 g' _" [( S) R5 V3 G
= 3X + 3 ( K^2 + K)
r+ c0 S* z( G+ |! U$ E = 3(X+ K^2 + K) which can be divided by 3
2 Y; c n9 Q& h. o3 s1 ^7 u
. ]. x8 ^4 i4 Z# J. vConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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1 y4 H/ f6 ?; R+ R[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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