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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 k9 ]) Y9 O3 u+ L4 H
! a" K6 a5 v+ s1 ?Proof: k* C" j: e" A" t v$ L
Let n >1 be an integer
+ W0 q4 @1 P# N. t( r" \' p# cBasis: (n=2)6 D1 \8 L4 e. G* ]* w/ ^
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 E3 y4 F( h' L: A" Y% {: a
& L# I. d7 X c. BInduction Hypothesis: Let K >=2 be integers, support that
8 b& }6 {1 i _" Q3 n, }% v K^3 – K can by divided by 3.& _$ m: {- Y% C- c7 A: l/ C
& K4 k( m8 @' d+ \1 l, C* R$ n' V
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3" U1 x: \. X) Z. p+ b' E
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
: `, S) [/ a/ D1 x$ J: d% EThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
% W4 n' X( K, b6 j6 e- H = K^3 + 3K^2 + 2K
% x4 g% R2 x$ ^4 b: W! \ = ( K^3 – K) + ( 3K^2 + 3K)7 }& `# J& Z% j+ b, e4 v- J
= ( K^3 – K) + 3 ( K^2 + K)" F# R1 Q6 \5 j# `) f# ~3 I' _
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>04 M, ?: |- q- h' Q
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)% [& c, e3 L) \4 a2 r
= 3X + 3 ( K^2 + K)/ \8 ~. i' [3 H$ Z$ ?
= 3(X+ K^2 + K) which can be divided by 3( a7 u9 Y% c. n: Y
" j5 F( `/ c7 A4 x. ^, f# o* O
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
" r! n8 e7 F3 v- ? f
- o8 O8 W$ ~9 g+ p8 `[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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