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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)
?5 O3 k+ @" ~- y
% v& E1 q- M3 q) b7 l/ H! XProof:
3 S1 C0 S. i5 A+ k3 ~* oLet n >1 be an integer
# m3 Q8 o R4 F+ `Basis: (n=2)2 O$ x. f( I+ V4 D1 U
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
q1 `0 V8 J5 o9 V# ?! ^0 T K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 30 @% n l9 k/ I. A" X& J* s
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem% y6 R* _/ F& E8 B' _* M
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)# I6 _: m8 U7 g- a8 x
= K^3 + 3K^2 + 2K
9 O3 w$ y; }3 W- k5 {1 t = ( K^3 – K) + ( 3K^2 + 3K)% k1 D8 M& X, u1 Q! a. H
= ( K^3 – K) + 3 ( K^2 + K)! Q* b# h+ v! E+ |+ l* X# A
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>03 ^/ r3 v6 x, t& N) Q3 T$ ]
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
# w6 o) p1 _$ ^4 k3 u = 3X + 3 ( K^2 + K)
" z# O! ?: H/ @1 V. y9 s = 3(X+ K^2 + K) which can be divided by 3
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.8 a3 l# m* E2 B9 E0 x
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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