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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: 7 \9 [) i6 A" Z2 o4 [
Let n >1 be an integer ; J8 L' }! g) B1 e' e" o
Basis: (n=2)0 W% v) Y9 K2 w
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3+ `$ M% j6 r9 P8 L4 s
9 {, b* N' e6 S0 X0 {, p; X& c! g
Induction Hypothesis: Let K >=2 be integers, support that0 m, P( k7 z" }( s' u
K^3 – K can by divided by 3.4 ?5 s9 O3 \6 |7 {: R
; l7 E$ P0 C6 @Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
; m, f2 r/ R2 s% u+ q. |since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem8 f6 ?" R! L3 ]6 b
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ T% [5 |% F* _# Y) d& u( u" o = K^3 + 3K^2 + 2K
4 F/ ~9 _5 u0 f) y7 f = ( K^3 – K) + ( 3K^2 + 3K)8 j) F+ x& E: U( V; a( ~% H
= ( K^3 – K) + 3 ( K^2 + K)6 z0 d8 u; q& d
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>09 P }6 \& ~& P, E: h4 C' y& {9 b" U
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
' x, N8 k9 [% i) Y9 q8 G = 3X + 3 ( K^2 + K)" d1 q5 d4 w/ p" y" ~& Q( |+ M
= 3(X+ K^2 + K) which can be divided by 3
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/ g2 H* ?2 J5 G6 Z8 J% `Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.7 s6 m1 R& W2 B/ q3 r( V
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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