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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)5 ]" \- f/ K' ^( e- s' M
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Proof:
5 g% v9 d2 O$ D7 q0 tLet n >1 be an integer
- W( l3 J6 W6 |+ bBasis: (n=2)0 X8 u" Y# a r) {/ I7 N1 W W
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3& q5 Y1 ?: f/ T) b/ g
$ x" x5 F6 |5 y0 ?Induction Hypothesis: Let K >=2 be integers, support that
% Y% ^) }' }3 }; t$ h. v K^3 – K can by divided by 3., _) P1 i; h! z% R( _
# ]6 C5 b" E" G$ RNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 32 ]& M$ B0 k3 r% z9 X3 x
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem/ ~1 N i7 _6 p' q" a
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1); ?3 n- d6 p7 {
= K^3 + 3K^2 + 2K
, `: G; @" V; o, C' }2 |) j = ( K^3 – K) + ( 3K^2 + 3K)
6 b) Q, {+ f4 t = ( K^3 – K) + 3 ( K^2 + K)
0 e0 o: O' |/ Q$ Pby 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
4 l+ k8 |6 W) s' S3 ISo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
. g3 B1 s8 s; o. J) Y1 H = 3X + 3 ( K^2 + K)) E3 u9 T+ `* ?- P
= 3(X+ K^2 + K) which can be divided by 3
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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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! K( F5 A, Y+ a7 _7 @# `, c* B/ s[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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