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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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[7 R: O# i% j! ~! `) G" eProof: 5 l8 g6 U% v, f, l
Let n >1 be an integer
- A0 r' g. o, ?2 W9 c3 `8 `Basis: (n=2)
0 u+ m) Y: f1 g( X ?9 F 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 U" u. ?( u6 V# C3 S( P+ l& e
2 v3 p) J3 j7 z) | ^6 J+ KInduction Hypothesis: Let K >=2 be integers, support that9 D7 g$ J9 ^1 J# a
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 3+ ^2 l. J5 P. I" D
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
8 r, h {! K) {* ^. zThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
( p# n! U) K# d5 Y' i = K^3 + 3K^2 + 2K3 ]8 e6 Q1 a7 Y. C0 ~ H
= ( K^3 – K) + ( 3K^2 + 3K)5 `; O" j; f" u7 n" g
= ( K^3 – K) + 3 ( K^2 + K)
2 S( i6 U( X, @0 {- K/ w2 @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>0
/ f2 k! h/ j- H0 X: qSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
$ a" V+ t3 V4 H- e M* B4 g = 3X + 3 ( K^2 + K)
/ l) M" r+ ?) j, H4 h6 o' a = 3(X+ K^2 + K) which can be divided by 3" f3 d" X# ]" X* g( N% J2 U
j: ?! i: e# e. B7 i3 zConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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0 s$ s% ]6 X, i8 r[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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