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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)# G5 J! U" z; O6 u2 H
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Proof:
- i" ^! d% ` d: q8 n( ]Let n >1 be an integer - q8 t$ m4 l/ k) [
Basis: (n=2)# z J; P+ O2 E: I6 U4 c( h h
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
, s5 O! X4 Y* U2 z2 c6 X }: B) [
Induction Hypothesis: Let K >=2 be integers, support that
# }8 \4 i8 n% n* t K^3 – K can by divided by 3.3 x5 M ]* O9 u* N3 l+ |) J
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
6 W0 ?/ T7 @( G& Q1 vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem3 m% }" i+ m, P |) Z) v3 E
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)4 p# V/ j( I; Z
= K^3 + 3K^2 + 2K
: w& h. Q/ Y: B = ( K^3 – K) + ( 3K^2 + 3K), c5 [7 U ~8 g' b1 `7 @4 s: g
= ( K^3 – K) + 3 ( K^2 + K)
) r- D F' v% f9 e- d4 yby 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
( U' e ]; Y- E9 ?6 `+ q% {2 I2 QSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K): ^( o$ k( |- M; k! p' F; p
= 3X + 3 ( K^2 + K)
) C5 {' d g# N! f = 3(X+ K^2 + K) which can be divided by 38 M+ Q& \! M3 }- L) m( p' k6 b
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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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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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