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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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& Y2 ` y& ^8 a; y" o( ^( PProof:
" x3 i$ J1 c( P, y4 fLet n >1 be an integer
7 W1 T1 r5 h. e; N& CBasis: (n=2)# W0 V# t$ l$ W3 W
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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6 |. f; ?; M9 s0 nInduction Hypothesis: Let K >=2 be integers, support that
% Y1 ]- q( N5 E5 u9 \7 c* E K^3 – K can by divided by 3.; j. f, S7 M1 Z/ }9 E K
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
! j: {. R8 x6 E! a4 K) osince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, ] _/ C1 `7 a o( R$ L& D
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
' E1 _% ^9 R# |* f j = K^3 + 3K^2 + 2K& c0 n$ x. t/ _: ~
= ( K^3 – K) + ( 3K^2 + 3K)3 Y5 c* h& A3 Z5 b3 \5 [& ]
= ( K^3 – K) + 3 ( K^2 + K), V' h% {& B5 ^* Z
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
7 f. X# j6 m- t8 l. y, xSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)( Z- j) E8 p: u$ F6 K
= 3X + 3 ( K^2 + K): D, J/ m; }+ A/ M5 {/ Q Q {
= 3(X+ K^2 + K) which can be divided by 3
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# c( r: M1 w8 ~# f3 x! R/ cConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.2 S. i6 Y6 D- {, ^0 R6 u8 ^
8 k4 S' O1 \" Z) y5 s[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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