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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)) f. X( T4 f" @' N& a6 a
9 P! `( j# _5 u- OProof: 4 e8 J5 Y- R' z* ?
Let n >1 be an integer ' u* s: q4 s+ t2 ~5 x
Basis: (n=2)
, h* D: p1 H! T( i* O# F 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3* ^/ D; Z' {# {' f4 e
7 W: u2 L4 b) u0 A/ n4 @Induction Hypothesis: Let K >=2 be integers, support that. c, N4 X# G, r! |7 k+ }6 `/ N
K^3 – K can by divided by 3.! A, K+ U( l/ m w2 t
' P# f. r2 E# q3 E4 G
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
$ N! t' S1 A4 e; isince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
D8 u9 E+ |! [ p1 d9 ~Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
+ q$ d3 ~7 g0 m+ W = K^3 + 3K^2 + 2K5 j8 U+ w- u& i% v- W
= ( K^3 – K) + ( 3K^2 + 3K)
" o& g b: ~- w( t" g X = ( K^3 – K) + 3 ( K^2 + K)
' @; W% I$ F3 U; `+ gby 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
) ?1 z6 i8 O: e5 J F% WSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) [7 Y& F) D! f& u- m = 3X + 3 ( K^2 + K)" M+ Z% P' Y7 \$ k# [+ p
= 3(X+ K^2 + K) which can be divided by 3
% w" a2 D+ F: `3 u3 \$ |$ M2 z" \/ }0 M6 l
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.4 |5 Q' g) |3 b. T2 ]
0 l5 f4 A6 J5 d) ?9 {# D W) ]
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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