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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) @, L! F( U. F" C
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Proof:
$ G, ?$ }! T0 b, f3 G; c! `: j0 W1 P0 oLet n >1 be an integer : B6 r3 o5 Z9 }8 h9 a R
Basis: (n=2)
1 A# }6 r3 B# p+ q! G) r* V# q5 W 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3+ r" u: e7 s; `* i) G4 b
% q8 Z- _6 j3 {& W" b/ J( rInduction Hypothesis: Let K >=2 be integers, support that
5 I( h4 k* y* b; u K^3 – K can by divided by 3.# m8 i2 k9 u7 X+ C) Z3 j$ ]: D
6 K: ^' F: K- |- y WNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3* F* d: G5 G! p; [
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem6 z7 S# o# ]0 J1 {& H7 l
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)! G& S2 x' h g8 k+ H+ e
= K^3 + 3K^2 + 2K
) j/ Q" J& T+ y9 a, V: ~4 w1 \% I = ( K^3 – K) + ( 3K^2 + 3K)( O7 ~% s6 M7 w
= ( K^3 – K) + 3 ( K^2 + K)& s+ K% }/ r) l( V
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>0
( k' u) j, m3 xSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)! g! d0 ^1 r) o* K
= 3X + 3 ( K^2 + K)
" Q* ^! x" e7 @. F; P, z = 3(X+ K^2 + K) which can be divided by 3
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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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b' A: p3 }7 k( F( Q0 B[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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