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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 D; M/ Z9 Q" UProof: ( `1 x( X$ W, o K* b& M
Let n >1 be an integer
/ ]3 n6 S! G3 M! KBasis: (n=2)9 E, Y2 W, d! s9 I
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 37 G7 ~, u) C; O8 I9 |7 X
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Induction Hypothesis: Let K >=2 be integers, support that
: h+ |/ @2 Z* u: `3 ?& r9 ]/ V K^3 – K can by divided by 3." M: d1 i: ~ U/ ?7 z8 f
/ j/ W" ?1 \9 W2 d, c+ nNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
: c. {6 i; c ?" Esince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
: N C) ~! k* K* |Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)8 ]1 r* A8 O) Z4 ?2 y2 J
= K^3 + 3K^2 + 2K
8 Z Q- k$ ^, M9 ^* w* i = ( K^3 – K) + ( 3K^2 + 3K)# S+ [; ?8 n' [. Z1 K" G
= ( K^3 – K) + 3 ( K^2 + K)
' s5 y; R$ g0 J: v& X8 Dby 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
6 s, m$ Z) q0 b, HSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)0 l4 D2 T5 |5 m) b+ o: S2 Q6 g
= 3X + 3 ( K^2 + K)
! p6 n, b+ `% _( |0 l5 V = 3(X+ K^2 + K) which can be divided by 32 p# |2 `- r K) f6 J7 ~
3 h+ l3 l: a; m. T( R5 iConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.# f% Z- e; M8 S4 X. S: C8 y
; q5 c0 u6 T$ O5 k) M1 a[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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