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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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5 j; E. E, [" ^2 z5 RProof:
' ]& O5 L, O# z0 Z0 I& nLet n >1 be an integer
' L' c3 h6 Z. Q1 HBasis: (n=2)
- U, s: ?+ c5 } 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
- ?3 c6 t4 c" m8 h% O! t
( o& I+ |0 Q6 {8 W4 h7 aInduction Hypothesis: Let K >=2 be integers, support that
/ [* ?$ Z0 [; ?2 _; H" G" d9 w# Q K^3 – K can by divided by 3., u3 H& v. X5 b' `
( T# b3 X" j% @6 O# }
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
$ R) `! u. i) n0 i) T+ Z: g, t B2 Asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
) H3 `1 {* E$ w; e. J; |4 EThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
% l! k. \8 ?! b, P" Z, g = K^3 + 3K^2 + 2K
$ s4 r' T0 V# x: C2 P; G = ( K^3 – K) + ( 3K^2 + 3K)
, Y0 U$ G! l7 c% I) q = ( K^3 – K) + 3 ( K^2 + K)) r( N7 H7 h) M0 C
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" y: f L8 o& |& t+ t6 V
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
# j' J5 U0 u& A( z2 {, v = 3X + 3 ( K^2 + K)
( I* I- m# a; e3 v/ Y e = 3(X+ K^2 + K) which can be divided by 3 ?( q1 @, ^) N. B& x3 X& o
; [" [1 r( S) C6 l( B1 fConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.4 C2 Q _) f8 l0 I# ]
1 T8 |# Y/ K/ ~! ?" @8 ?( X[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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