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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)7 F Q. T/ [ s3 u# u- W
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Proof: % _% k4 }- S0 q0 e8 ~7 u
Let n >1 be an integer
; j$ d7 g! t ZBasis: (n=2)
+ u Q: x9 i- R( g, e' P 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3% {; n# S% z, F. \2 K I
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Induction Hypothesis: Let K >=2 be integers, support that
# P/ w, v7 N U5 g. y8 g K^3 – K can by divided by 3.
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; f' y. M" T: PNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
8 t7 o1 _0 r, U% `& z, b6 l* Asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
$ Z; M. M" i6 Y1 c2 e/ P) R" l. TThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)( X9 [, d, A7 P& ~- b
= K^3 + 3K^2 + 2K
4 N1 I# H6 t- r! p = ( K^3 – K) + ( 3K^2 + 3K)) G& p3 Z( j5 ^4 k3 O- Z0 z2 v
= ( K^3 – K) + 3 ( K^2 + K): _% C3 c7 \# q7 s/ M/ J) G
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>04 u, M. [, k: j; Z2 P" R# ?0 ?
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)9 c. ], C7 a8 e" b% x
= 3X + 3 ( K^2 + K)
9 ]* k5 \& }) x) [ = 3(X+ K^2 + K) which can be divided by 3, e3 ~: s6 K l" b) u$ v9 H
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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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