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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)
& y1 j3 U! \ U' t; s/ v3 k) k; {" H- C! @
Proof: 1 c1 ~4 a% {5 U' M' |% M! b1 c
Let n >1 be an integer
% [8 D5 v4 y0 T1 Y7 {9 LBasis: (n=2)
/ a, c9 Q/ i" w$ b b 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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, g- S( a2 Y% `9 v& ~2 `Induction Hypothesis: Let K >=2 be integers, support that5 P, \ c* \- s1 F% b5 Z, L# B& s
K^3 – K can by divided by 3." }3 ^3 ?% l% _" ~3 \! L
) C2 |" q; u( \) P/ @Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3) [) a' z3 g' R$ \9 P, N
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
- ^8 Z6 @- i# N2 n NThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1): \5 w; j. e; E5 U) w3 s
= K^3 + 3K^2 + 2K- X" R/ `1 U+ _- |6 S/ X( m
= ( K^3 – K) + ( 3K^2 + 3K)
% e2 ~% W) Q1 ~( x2 o = ( K^3 – K) + 3 ( K^2 + K)
& @# V. Y0 v9 |: ^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
2 B: _8 s* ? aSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
. \, U5 I. I9 K n( ?2 t" o = 3X + 3 ( K^2 + K)
q8 q* F! v& y+ O8 \8 |4 f = 3(X+ K^2 + K) which can be divided by 3. z) y4 a! p# y( u! ^8 A$ G, `4 U
! V) ?" x- D/ o" BConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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_! a7 E7 b+ c8 J9 d* Y" } w[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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