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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)
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Proof:
- D# [" u; q% l7 j( `Let n >1 be an integer 0 ]2 N) E0 }+ [6 I
Basis: (n=2) P3 I3 y2 w/ R" [4 S$ X
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3! _% l- H9 e/ ?# F1 ]5 W
/ D' Y* _6 q1 r/ L( ?8 n ]Induction Hypothesis: Let K >=2 be integers, support that
, O( x! j; c& W- a; x% W; B K^3 – K can by divided by 3.8 r3 k3 x" N9 N, Y4 L: ^
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ R& z$ o( \) _3 ]! C9 Ssince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
9 M1 |1 T; x6 D' ~7 IThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1), H5 @4 O" v% d4 M) }
= K^3 + 3K^2 + 2K* _" D4 k2 [! {- \0 P0 w
= ( K^3 – K) + ( 3K^2 + 3K)* A* f- U3 @% D/ f' t7 Z+ S
= ( K^3 – K) + 3 ( K^2 + K)5 G1 `" I% l H) `$ p* [
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. `) H/ m, k! G1 ?/ C7 |1 b
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)% D7 k; ^- s' X7 [+ g
= 3X + 3 ( K^2 + K)0 P: `4 d/ D" H' T; O9 x8 b
= 3(X+ K^2 + K) which can be divided by 38 G; w! C6 H/ 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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C. t6 @1 c# s( u. z[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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