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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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Proof: / m5 c9 r: Q$ b9 s
Let n >1 be an integer
# I/ N2 {( n* x+ N/ m- w+ ?- HBasis: (n=2)! x* I8 Z1 ~( Q5 R
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 [4 i8 e5 a1 q$ k7 G& F
) a% R8 o' q; O! U% O" b4 S" HInduction Hypothesis: Let K >=2 be integers, support that- y! w- a! M! z p& U
K^3 – K can by divided by 3.9 B j' _* | V
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3: z# ^9 q* C! d/ b0 x: W N
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem) i% O, \% Z8 y% L/ ]
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
4 T/ P) d( P8 O( S = K^3 + 3K^2 + 2K) G+ o0 w; S6 a6 H
= ( K^3 – K) + ( 3K^2 + 3K)
/ n- ?/ `) A) W0 J$ ]( [' b i3 D = ( K^3 – K) + 3 ( K^2 + K)
3 c* X& T! H3 G0 B( iby 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
0 j5 }5 N! V8 eSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)4 P! b- i0 t- t& y+ t/ l% D/ ]
= 3X + 3 ( K^2 + K)
6 b4 @+ L1 ?2 s( b$ Q o = 3(X+ K^2 + K) which can be divided by 34 a+ r S* ~, B3 |) k0 a7 K1 a/ b
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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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