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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:
6 N8 l2 l* r0 }Let n >1 be an integer , v* k- P) Z O, Y$ S
Basis: (n=2)
8 _; X# W4 u3 U0 V# j \" M3 U 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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& A8 W7 b9 D" f7 A& {! UInduction Hypothesis: Let K >=2 be integers, support that
4 p. L$ Q. l6 n& ^2 M: D K^3 – K can by divided by 3.( W `4 K5 U! k0 G1 M4 e
+ p# q" H% y5 e3 ?: N/ _& V) l- |Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
) P) T3 }8 v/ N: a* U [since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem ?4 ?4 w( t; t& g( ^) H5 u1 L* B& D
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
1 ?) s* g& T0 ] = K^3 + 3K^2 + 2K% P% s' w* f( b7 t. w8 j B
= ( K^3 – K) + ( 3K^2 + 3K)
/ T! P+ l+ x- r! M6 C = ( K^3 – K) + 3 ( K^2 + K)! _7 x0 ?: e5 Q( N5 Q" K( z
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
9 m6 t( z5 A* w. |So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
0 u5 m; X) Z) b. [( q$ ^ = 3X + 3 ( K^2 + K)
+ _9 ?, ?7 F, E$ W$ J% {1 e = 3(X+ K^2 + K) which can be divided by 3. ]2 U5 d7 h. h; v* R
7 Z3 x" ^- B" \3 T/ E+ |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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