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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: # m. b8 L! I4 S2 u) [- U$ W3 g
Let n >1 be an integer
1 F' x4 A! z w1 G% n, wBasis: (n=2)9 \4 f2 ]; f: `7 T0 D0 z
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3, r7 b X2 W2 |
, h. R O1 v2 dInduction Hypothesis: Let K >=2 be integers, support that& ?, @! ~5 m2 o/ u, H
K^3 – K can by divided by 3.$ R5 ~5 S* @9 l- ~- {
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 38 M% K% [$ z) L5 J3 y7 M
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem" t( {0 {1 {. F) m/ G& G( V9 |
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)- X4 y; G) _4 k) l, K" Q5 j
= K^3 + 3K^2 + 2K' e' ?1 b" g7 Y) V
= ( K^3 – K) + ( 3K^2 + 3K)$ ]3 L. E# N5 }! a) G& o0 b
= ( K^3 – K) + 3 ( K^2 + K)+ ]5 r \2 {" h
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
3 i; N J& c0 `% F- oSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)# e. E, t2 w/ |6 L" {) u$ n/ p5 p
= 3X + 3 ( K^2 + K)2 [; e+ {8 Z5 n
= 3(X+ K^2 + K) which can be divided by 3) R+ l9 l1 K" ?1 e& e# D8 \& w
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.: x O7 R8 v0 \& |; q J
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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