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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)5 H/ g* a, q( ]3 O$ K- g. p
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Proof: . M. t- t) I3 Z' R" x: g
Let n >1 be an integer
$ H! C" j1 q1 ]( g" @8 NBasis: (n=2)
! w/ [' b6 \0 H/ J A 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# o5 \* Y/ _" y' t0 f% K+ T- L
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Induction Hypothesis: Let K >=2 be integers, support that' G W, {2 g3 S, }+ A2 d
K^3 – K can by divided by 3.2 E! b- Y+ y# V- x- Q
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
& b) B& b- H( s! O. V& csince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem9 m3 o" o. N8 ]7 u- a: M
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)7 {% r/ ?8 V/ C' `0 `! r
= K^3 + 3K^2 + 2K
7 U4 t9 N4 H& @7 a. x = ( K^3 – K) + ( 3K^2 + 3K)/ {7 h1 v- L O9 S' \3 g8 x
= ( K^3 – K) + 3 ( K^2 + K)
2 y1 K4 r9 a" A$ j" w0 {4 Vby 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& k& T2 x$ }" U5 l( t
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
2 d) v# e l* h' S9 M, R, ?/ y = 3X + 3 ( K^2 + K)/ [2 n* B$ n, [3 L; v
= 3(X+ K^2 + K) which can be divided by 3) U5 \2 g! F' u
% o( [1 t, [6 e1 v/ ~" VConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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7 D3 W: \/ l" ?" M1 z[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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