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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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4 J, j4 B9 }6 e- \) G. h* k0 wProof:
( S% ~7 I i0 kLet n >1 be an integer 2 M7 m& Q% C" y% ?0 }8 |. k2 O
Basis: (n=2)' R$ w, r& y( I) L% i5 O O4 q) g* _
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 w/ L5 D# E# v" P# w" _ ^
+ M2 n' u4 Z( n6 \: eInduction Hypothesis: Let K >=2 be integers, support that
! B, c1 {6 e' q1 f 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" ^) w# e* x: d. _
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
% q+ D3 ?5 u5 p3 RThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)1 o* o: i: d' k# O' N7 j
= K^3 + 3K^2 + 2K
' ^- ^- L$ H' H; Y2 d6 } = ( K^3 – K) + ( 3K^2 + 3K)1 Y# N, s( T# h! L& V+ ^
= ( K^3 – K) + 3 ( K^2 + K)6 G: G5 ]' [2 e% ]! 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
- P/ L- ]$ q5 V; D9 ISo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
! R5 r5 h: U/ B0 Q4 o1 O = 3X + 3 ( K^2 + K)& h+ h3 I$ a8 R' j' c0 C
= 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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) w2 q- P. c& H[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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