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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): U5 d, h) u. E6 l8 v* U' V
( h7 l0 U9 Z) o6 ]1 R2 n
Proof:
9 V& o- u7 c% h- g# W2 xLet n >1 be an integer
* G/ q5 @: k( P# kBasis: (n=2)
+ k+ [! d+ U% j: \( y: h& P5 V 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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& s- {# x1 v7 h/ m* uInduction Hypothesis: Let K >=2 be integers, support that. N7 K& J$ {$ ?5 W
K^3 – K can by divided by 3.
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; T# V# S g! Z( ^/ @) k# I% [Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, l8 W) i9 x1 B9 V- Nsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem4 ] _) J/ m; b* F ^$ |. L4 M
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
0 y7 M4 p/ G: d2 F = K^3 + 3K^2 + 2K2 U! X2 c9 f* w3 b" r0 m. m
= ( K^3 – K) + ( 3K^2 + 3K)9 W# L0 m( t3 a: o5 M
= ( K^3 – K) + 3 ( K^2 + K) X; }0 u* F. O+ 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
! F$ n' M# F5 j4 a1 j" q, E( l6 ySo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ {4 v- W& S! _; O* `4 C/ q = 3X + 3 ( K^2 + K)/ z) z2 u- @- W X) n) g
= 3(X+ K^2 + K) which can be divided by 31 O( p* n: g0 } h& X) V w# K
, `0 A+ C$ C/ ] ~+ {6 G$ F- x! Y1 v
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.+ s& f8 U) R% F7 h$ ^
! }- k; z0 p$ M/ D8 k8 O/ I2 S4 E$ s1 ^[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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