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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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; p( v& F- q' i7 }, M" OProof:
+ b0 X7 R5 T; }7 @ S! H4 ILet n >1 be an integer 9 P6 a! o- @! ^; C! e7 S7 E
Basis: (n=2)+ a8 ? S4 @8 x+ K
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 37 P7 L# E; n3 N" w* O) B4 X! ~- r
( E0 ~0 R" x3 |& _# m. XInduction Hypothesis: Let K >=2 be integers, support that
' @; A9 h# Q' N 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# o O" t* o: E9 y
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem5 h. `% S" C% x0 t3 U# Y5 t7 g9 B8 w
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
7 l+ N) S1 a5 K i; e8 |6 R5 l8 H = K^3 + 3K^2 + 2K
+ I* t3 c2 y+ e4 d. W% P4 { `( `7 ] = ( K^3 – K) + ( 3K^2 + 3K)
3 }+ s; y; L. B0 k = ( K^3 – K) + 3 ( K^2 + K)+ P3 T1 u; h8 ^5 y! @1 [
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& s* o) u! I1 ~' O& N `+ J1 ?
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
& d& h% ?, ^( l# {2 K, h% ^ = 3X + 3 ( K^2 + K)
( u! b- s2 o, E0 x- G+ t3 _, Y( U- E = 3(X+ K^2 + K) which can be divided by 3
$ N1 e$ K3 f m* @/ P$ L# T) q) T6 r+ o- q0 Q: F
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.3 k& S! R% J$ ?4 R: y$ k
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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