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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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$ i# ^1 T9 h! ^4 u( d; oProof: ! N1 ?( x4 Z5 ?# c3 ]
Let n >1 be an integer
& d3 N$ d3 ~ x6 }4 L$ |% hBasis: (n=2)* [# |6 C9 M/ Z& [) r3 T% q
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 32 a9 J+ `) M) m" Z! ^3 O3 S
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Induction Hypothesis: Let K >=2 be integers, support that
& q% }6 P. c9 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 3) c7 L/ C) O$ ~8 n: H/ g, g- b
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem5 O( L# ], z- v: t- y5 W$ Q
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
! Z- x* s6 s0 q- r9 a; P2 T8 u = K^3 + 3K^2 + 2K
! h( o4 ?' m/ K' C2 K* e- U5 J4 V' N = ( K^3 – K) + ( 3K^2 + 3K). W: U- [, ^6 r
= ( K^3 – K) + 3 ( K^2 + K)
- i7 X0 ] D; x1 c* fby 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- l3 n8 G1 j+ R6 @! i7 b
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
0 }8 b# t; n; d* z2 D( J = 3X + 3 ( K^2 + K)- k1 `. g9 r& W) E" m! \: C
= 3(X+ K^2 + K) which can be divided by 31 d0 m5 T. ^% j8 _
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.. r4 u: G. \, R. C5 ~8 @( ]9 s* K2 N$ R
( @1 @% b2 f" j$ B- J[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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