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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)! f4 X2 Y4 T0 [3 t
* w4 |& I4 S! D3 U$ q4 {7 C7 lProof:
6 C2 x. N/ R7 u8 h& ZLet n >1 be an integer ( q# |& E$ V0 @
Basis: (n=2)1 s* p* ~4 o; D. [; u$ W' k+ h( j
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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+ m8 Y6 r$ S2 v( F- I( _Induction Hypothesis: Let K >=2 be integers, support that
0 i/ F. W9 |# c6 J B# d: k K^3 – K can by divided by 3.1 W+ I. a% Y+ p; M7 |: J
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
5 s1 s3 j+ _9 S' V R7 Bsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
: y; m8 F4 ^$ @Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)9 r: A1 y+ [# ]
= K^3 + 3K^2 + 2K( |; v- m7 D6 i8 V M, y
= ( K^3 – K) + ( 3K^2 + 3K)
9 T" ^3 ]3 @9 B, `2 n$ G6 [: L& n = ( K^3 – K) + 3 ( K^2 + K)! Y+ J4 M; ?' z! L
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 w" ^2 o+ W2 t9 e
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)- D2 a2 ?! k% r3 t
= 3X + 3 ( K^2 + K)
1 Z1 {2 J( D) {5 ?) \ = 3(X+ K^2 + K) which can be divided by 3
0 A; }0 v- w% I3 `& C/ i% e, m5 c7 f2 t; f
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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: c. g* S% c9 m8 G0 q. d& b* K+ `[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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