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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)1 ?! X w% C5 Z$ D
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Proof: . b1 x( A. x8 h% t
Let n >1 be an integer & m" K( B. I. X- L0 x
Basis: (n=2)
; u8 Y' Q2 ^' x+ C 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3+ Z1 Y2 u) l& X6 m7 e: [2 k
1 z# \$ j8 |0 d7 K' @' M: ^Induction Hypothesis: Let K >=2 be integers, support that; p3 b. I( u4 j1 L' R" S7 N$ r
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 39 U9 F' ] Y' m5 B1 v
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem* k0 v. k. Y& k) F- N) l
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)4 h% l O" u4 r4 M
= K^3 + 3K^2 + 2K( \* E. S# u% U4 R# N* F2 s
= ( K^3 – K) + ( 3K^2 + 3K)
% {) W$ ~4 M- l6 A h = ( K^3 – K) + 3 ( K^2 + K)
( w2 ~. L" C! ]1 M. [& E4 i4 cby 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
5 X) v& T7 Z! cSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K) S d T2 V9 t( l9 G
= 3X + 3 ( K^2 + K)9 V1 l. o% C$ \3 D4 Z
= 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 >1.7 `% [1 p5 T% z, r
" h8 A4 e6 Y/ O5 `[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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