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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)
% |* A) E w, j; v
8 L! n# O) U( i2 V! O9 ?1 O; D2 t" yProof:
4 ], U" ]6 n$ ?0 H# y* x1 v; LLet n >1 be an integer + l0 [+ J' i: D, s! d. u, ?
Basis: (n=2)
1 e# Z2 j1 A' j: _ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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" J9 d0 I6 }: n9 T7 H# d% @, \# I8 @Induction Hypothesis: Let K >=2 be integers, support that
6 w7 V7 T% @: H+ x$ ?/ ^, u; D K^3 – K can by divided by 3.9 D% n3 B. {% A: k0 y5 @
4 I W, J$ U" V2 X: I& H8 A; RNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
- f+ C" G0 {# |4 ~since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
# P+ ^* k+ h6 ~, Z+ `9 hThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1), i Z9 K# }7 V! e! r; M* O" U3 m
= K^3 + 3K^2 + 2K4 U! u7 w/ O1 o& f/ l
= ( K^3 – K) + ( 3K^2 + 3K)
$ }2 m2 U, J0 d) B+ v; S) [( _* o1 ] = ( K^3 – K) + 3 ( K^2 + K)* a9 @" B/ G7 N; S
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
( S7 l/ V7 f! b6 q* jSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
8 Z A& v! T. k/ T8 g1 w) I = 3X + 3 ( K^2 + K)
% y) c2 u. h4 c) h' T = 3(X+ K^2 + K) which can be divided by 38 Q, r$ G7 U( ?' C% d9 t2 V
7 K# p, Q* C! k9 A
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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