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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)& i& s+ F+ q d+ A: k J+ Q' P2 B
. K+ E; l8 V4 j, nProof:
[: I$ q. \1 Z) [( a, wLet n >1 be an integer
" \' L$ P( Y! @) u7 {; _Basis: (n=2), e- N7 u7 J. I' k
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 37 Q& L. Y& w. Q' J8 j( X, S
L( [( |8 }. ?! j
Induction Hypothesis: Let K >=2 be integers, support that
4 o* e9 D7 |- C6 v8 s+ t K^3 – K can by divided by 3.
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2 ]0 D8 v' P4 n# Z7 {+ xNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
! D- z$ |; P( P' o! Isince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem. t, R7 \) X$ v- @8 j7 ^
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)) ~1 [. F5 A# r. W5 n$ x
= K^3 + 3K^2 + 2K
7 m9 w- [" S6 N = ( K^3 – K) + ( 3K^2 + 3K)/ [: {% y' @7 S6 B% ?( x
= ( K^3 – K) + 3 ( K^2 + K)
: f* A" `4 R. o& R* Lby 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
' W0 J. @0 @( e. t0 {5 ASo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 \' {7 i2 {; o- q; Y = 3X + 3 ( K^2 + K)1 I |; E, P8 k/ d' j" w; }; L
= 3(X+ K^2 + K) which can be divided by 3
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. _4 A j9 k' B- ^+ `9 wConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1. w# j6 _, h& C
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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