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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)/ S/ ]8 o" r- H
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Proof:
x' E- a; r% {8 Z6 s. V7 |Let n >1 be an integer 1 q& A+ ~" K" ?+ F3 q7 q9 F, @
Basis: (n=2)
( I" Q3 K' Q0 K 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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: \9 r+ Y! |; P3 h- a/ @Induction Hypothesis: Let K >=2 be integers, support that
* \3 z+ z2 c7 ^5 ^$ Z0 [9 o# Z0 B5 B: ? K^3 – K can by divided by 3.1 J$ _9 q* g+ z
6 O' r. j# _! C V$ L4 g9 h* eNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3' v. T7 }# e. _0 T
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem3 A( ]7 ?0 l; m* W8 c' Z
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)5 J! w: Z% Q% V4 Z8 Z
= K^3 + 3K^2 + 2K( c0 Q( t8 q" [1 E4 o
= ( K^3 – K) + ( 3K^2 + 3K)# }7 \0 ?/ C. \3 s0 [; h
= ( K^3 – K) + 3 ( K^2 + K)/ X4 U3 t( S1 G# u' q6 i1 Y
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- M( @8 _$ \7 U! Q- M
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
2 @0 _& F4 j8 q: U- @- K# w = 3X + 3 ( K^2 + K)0 T) ]) Y" ~5 B; u/ u: w; J
= 3(X+ K^2 + K) which can be divided by 37 A% S% d5 U0 S7 t
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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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