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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# G9 P1 _' [( t; x5 sProof:
: ^5 Y: Q9 j# q% G/ E4 xLet n >1 be an integer
z4 T2 I. t1 z0 y: O2 ]Basis: (n=2)& Z/ Z6 L/ k+ V1 o/ i8 h8 P& k
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 38 j+ t |, a- C" |- m5 W+ ~! L
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Induction Hypothesis: Let K >=2 be integers, support that
% u0 C9 R4 |: V h K^3 – K can by divided by 3.& s% n! p4 ?2 v% {, b! I8 ^1 I
' T8 ^" x6 S4 w9 A0 x4 e: GNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3, s. v: f4 X8 A% \4 K$ X# Y1 V1 m
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem$ g6 z% c5 u4 P# s1 k, s* c' d: d3 }
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
' Q; W' ]% }5 A- [; c' w6 @0 t = K^3 + 3K^2 + 2K$ Y5 y, j7 Q Y7 `' Q" B
= ( K^3 – K) + ( 3K^2 + 3K)# H X' r1 K! H6 ~3 s+ B
= ( K^3 – K) + 3 ( K^2 + K)* D2 o* e X3 T2 B' o
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
! y5 m/ R5 }, x5 T4 F, @5 rSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)$ m' y/ @: Y. j5 p! G# q9 v
= 3X + 3 ( K^2 + K)
' w: P4 J6 d0 L! j7 G1 U3 u* n) Z = 3(X+ K^2 + K) which can be divided by 3
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2 S* u! s8 O. xConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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' x) q" F3 d8 k5 C5 _[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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