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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 ?6 C @$ M F6 O
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Proof:
9 K" Y& T2 J9 C+ h! O7 l l9 ]Let n >1 be an integer 9 @' l8 d7 L! [0 l4 k5 X; B P
Basis: (n=2)
) C; A5 s+ f) Z3 l0 R# d: l0 U 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
7 Z+ j* w! N: k" V' t! X: ?$ ]/ e9 O7 [! v v7 ~2 B
Induction Hypothesis: Let K >=2 be integers, support that
& l5 U& ?# |+ l# \6 p8 W6 U; n3 c" I K^3 – K can by divided by 3.
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0 r& ?4 s2 G x8 Y V" K( _Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3. N& m) p) M3 |( _' @! d
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem; E3 {' T) q+ Y) l C
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
3 J+ Y! p/ w. M1 S) K( L; N = K^3 + 3K^2 + 2K4 e# t- N8 `7 J
= ( K^3 – K) + ( 3K^2 + 3K)
, l- E3 c, A. n8 B = ( K^3 – K) + 3 ( K^2 + K)
$ A& p s* x9 z# p9 ~4 Uby 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- m# {+ \& S8 G0 L) ~$ X
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K), ?2 t4 n! c8 g, n. h
= 3X + 3 ( K^2 + K)
" m, @, f* s3 K0 b( }7 ? = 3(X+ K^2 + K) which can be divided by 3
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% c7 \4 w! M( _' Q5 D# P, R, [Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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