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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)' m" T4 t+ I. B; n6 i
3 T; s: f8 ~( q8 b
Proof:
. H& ]7 _$ j' @Let n >1 be an integer , t5 K: |" P5 W" ], R
Basis: (n=2)& H6 ]/ P6 E; ? ]/ P
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
& I$ k; p( [, C& G0 K! p$ I! d+ V* y
Induction Hypothesis: Let K >=2 be integers, support that
4 o0 Q1 N# ]! W7 Q- L K^3 – K can by divided by 3.
h- `4 l" J4 [( }; B* {' O
9 K6 m0 c- o. m; O+ g6 n8 f! JNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
3 H2 k. s% ^, T& ^since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
& K k% }+ t8 }4 k! p$ i, `Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)5 L: s( N; f8 j+ j
= K^3 + 3K^2 + 2K
) M& y3 j, k4 O+ W9 m% i* i = ( K^3 – K) + ( 3K^2 + 3K)2 R4 c {* F# A) A. I* C
= ( K^3 – K) + 3 ( K^2 + K)9 k( E* E- }& M0 b0 A* 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>0" q5 i; Y5 O& k& x* A6 }- _$ f
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
* u& G7 Y! m) g9 V/ R- m = 3X + 3 ( K^2 + K)& K, o/ S) ]0 k
= 3(X+ K^2 + K) which can be divided by 3; ~/ r% \ ]3 y( s: b
9 ^+ w4 T( A; d/ ]6 `. j' d/ O: ~Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1./ X+ N. X) A& w& p
' O6 v. X8 ~! |* V2 J! ~6 Y[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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