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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)8 b$ _: \# N# `2 E2 w6 P
8 c; f7 Y# D' U: ^. W
Proof:
h. N4 G8 |, a+ j3 S3 o2 f) M; [Let n >1 be an integer
" R* X" H9 N' S. k7 PBasis: (n=2)3 V. j) v$ Z7 s
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3, x. z. g: S3 e! N4 p
) a) b# ^9 i$ k" i$ d
Induction Hypothesis: Let K >=2 be integers, support that" p8 ^' Z4 A9 Q( d+ J
K^3 – K can by divided by 3.: ~0 T1 K* ]( g
0 n1 ]& F! T, BNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 36 D, @% k |8 G& j! v y
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem( v) t/ z F, I6 f# M" E
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)3 f9 I2 N# ]( n
= K^3 + 3K^2 + 2K
9 w3 k5 j& f" Z( u7 O/ N; @5 a# E7 v$ V = ( K^3 – K) + ( 3K^2 + 3K)
_ z2 U2 h( }0 E = ( K^3 – K) + 3 ( K^2 + K)+ K) G3 M* Q7 }, q1 N; C
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>01 [' w s& J" v1 X
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)$ F6 C1 @5 r) ]$ k
= 3X + 3 ( K^2 + K)
; }8 O: E+ W: H: s6 {8 ~8 O = 3(X+ K^2 + K) which can be divided by 30 F& m% |% c2 R# k' \
8 h* {, F# ^4 k
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.. H# m/ Q0 n. G1 }8 g' R
+ E8 [9 D* \/ ^# ^# Z[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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