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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)
7 u% U5 P! H) f+ N- G+ r+ x: Z! M1 T0 D: Z4 G- b. M4 V4 x
Proof:
5 g" F( \2 p5 h+ N2 ?2 R$ K! V2 i. sLet n >1 be an integer
, j% V& B3 k9 i. k/ `Basis: (n=2)! g9 y# E& m' L p, h2 j
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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$ ~# f. s" l' }1 H. K( tInduction Hypothesis: Let K >=2 be integers, support that
0 g! {, y: ]# q K^3 – K can by divided by 3. X" ]4 Z. X; G; ~6 c
5 t" X( M" u4 S0 RNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
2 I% B) ]$ S; Rsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
2 ^$ i0 u( T6 K& g- {2 y7 n* {Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)! v$ M' H& c8 R* J. E6 A' Y7 Q
= K^3 + 3K^2 + 2K
. x: V. F+ D8 x3 F Z9 v+ k+ B = ( K^3 – K) + ( 3K^2 + 3K)
& Z% L9 u6 x1 M( U+ d3 _% M = ( K^3 – K) + 3 ( K^2 + K)
( J3 _. L( q/ v3 Y: eby 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
( a# @( `/ w* n5 A0 S, A0 x `So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
8 u/ M [7 ~0 X: J, r3 q$ [ = 3X + 3 ( K^2 + K)
& k) L, W# \0 F; [ = 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 >1.1 A3 `% m4 ~# w2 K a
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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