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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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$ X! I3 T w1 l$ g+ ~* d1 rProof: 5 o6 e3 G+ y' D2 m$ C! ?
Let n >1 be an integer 7 I8 M5 m. S2 z# Q3 @
Basis: (n=2)
/ f5 V) B& t4 L6 s7 _. W 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3* x# E$ i! p, O" j6 |. c4 u
4 s! ?0 }( k) P* F6 H( mInduction Hypothesis: Let K >=2 be integers, support that
% N$ B! n9 |! K9 F$ r! u% H K^3 – K can by divided by 3.
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( h# v& j) } k. H' c* V4 QNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 C+ Q) |. s6 n# K* G3 R" I# S7 osince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, G: Y8 B' y# V" ^
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ P$ D5 j' c, X+ Q% {# T = K^3 + 3K^2 + 2K
) q8 \6 f8 z1 O( S = ( K^3 – K) + ( 3K^2 + 3K)
, \) @' x: H* q2 o = ( K^3 – K) + 3 ( K^2 + K)
1 n- n2 S, l8 }. R6 aby 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
+ y$ U' o4 V: M- y j0 {So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ b o' o! E4 ?/ A* t% F
= 3X + 3 ( K^2 + K)
$ \ f6 c! q' t& h = 3(X+ K^2 + K) which can be divided by 3" A6 x G5 R+ H6 A' ?4 U
; K- O% n, Z9 ^, g' mConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; W, t$ t3 N8 F! S& g
1 p: a4 z* `0 `8 ~2 ^[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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