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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), p+ _! z5 z' P/ z4 W4 S+ u
3 ^0 J3 \3 H0 n8 l1 X! fProof: ( Z5 C6 l4 k+ V1 _1 U+ f0 I2 j
Let n >1 be an integer . ]3 j# w8 K4 \$ C* z8 G
Basis: (n=2)
- c8 Z2 ? N; ?) |- n0 b 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 M6 U; ]7 y- Z1 U, g: ]
. y, }7 v4 {' s7 aInduction Hypothesis: Let K >=2 be integers, support that M, P" R" `+ f9 a0 N. ]" q' J9 q
K^3 – K can by divided by 3.
; f2 u* c' {2 h1 b; \
% R7 H# y$ }, [; i7 cNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
2 B/ W" ~) [: V! B& X Qsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem5 X7 i3 @ V+ l0 p# q
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, U* |" D6 s- i( J = K^3 + 3K^2 + 2K
- h' G7 Y. |' L% q/ ?1 t9 K1 { = ( K^3 – K) + ( 3K^2 + 3K)" A; J v8 l, s, M7 Q% b1 r
= ( K^3 – K) + 3 ( K^2 + K)9 [# k/ u9 z( w; N- E- s P3 L
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
5 u: i5 i' F" E+ V A, _/ I) ySo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
3 k7 B3 _" t+ K3 e. V# {2 T1 ? = 3X + 3 ( K^2 + K)
) `( C3 p( J u( V = 3(X+ K^2 + K) which can be divided by 3$ m% l2 L5 A/ W8 R, s" n3 F
. x+ R$ l0 q+ M7 ?: y. }$ H( h
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.# T- o# [+ I0 A" B3 x+ s v7 {* l$ Y
! E! F* r0 N/ \" B6 b) o. j
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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