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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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# Z6 s+ l9 ]7 z6 Y. r' BProof:
1 M- P0 Y! c+ ~2 v! TLet n >1 be an integer . W+ T2 S9 d1 {, r. b
Basis: (n=2); n, ^9 x( H0 {1 [4 c1 {
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 a! V b* W# m& t& z
' I N6 |9 T. qInduction Hypothesis: Let K >=2 be integers, support that, n# O; u: c! W6 ]% S
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 33 P1 v$ b! ]6 S7 i8 D: G8 p8 |. J' C! l
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem9 m, i' u6 l+ ] ?4 q3 d
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
% d' [7 I( n- a( \& s = K^3 + 3K^2 + 2K
7 f# T# _/ r0 C! r* g% O% |1 o& ` = ( K^3 – K) + ( 3K^2 + 3K)
. b' V% t4 V: |4 c* ]* n = ( K^3 – K) + 3 ( K^2 + K)1 v5 u* Y: I, N
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>04 E& ]- [1 c6 K3 W! A% ?% A
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) \7 S9 P" D8 G$ o = 3X + 3 ( K^2 + K)
4 o! V$ k/ ? F+ A/ M A = 3(X+ K^2 + K) which can be divided by 3+ J! C0 d7 J* c
0 C) j+ r& N1 V' v. rConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.- Y2 f% r- M+ Q* R2 _
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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