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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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Proof: 5 b( @* u8 k, T2 E
Let n >1 be an integer
0 M i( u+ }, Q* A% kBasis: (n=2): H# v5 ^0 T% I3 l# @# f# T. d' a
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 38 P* M, p( ? ^; `& ]
; u7 t5 Y4 m7 `4 OInduction Hypothesis: Let K >=2 be integers, support that
* d' R( i! a/ n K^3 – K can by divided by 3.. g1 X7 S8 B" P5 T; W
% f" ~6 q$ k* m6 @. q9 ^; W oNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 38 X" q6 r- r) `$ C
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
% s. d; L$ `- r, mThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
% y& F, {" J! i4 E$ F! S% R = K^3 + 3K^2 + 2K
& k9 J1 H( H# g: U* ^% n = ( K^3 – K) + ( 3K^2 + 3K)
- x8 \! _' M( C- r$ u! R: _ = ( K^3 – K) + 3 ( K^2 + K)
7 k4 a7 A6 ~0 P: L8 c' pby 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
+ c0 P S6 O* F0 m9 I8 t2 pSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
& Q1 i; X! L+ Q" q! F; l = 3X + 3 ( K^2 + K)2 Z2 X, e! X2 z2 v" p
= 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.& {0 B/ c2 r( p' F7 E* @, p% \
4 }: }- C& q& g3 X7 L: c* v$ v7 p[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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