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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)) b+ J7 p! ~$ p* `* ?
" `$ |+ t. Y8 mProof:
! _/ ]9 E/ H: ^- o1 d" z: yLet n >1 be an integer 8 B' s `0 x& z) M
Basis: (n=2)
( C2 i0 `, P3 ~$ `; C/ W1 d 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that: \! o& g k+ I. D
K^3 – K can by divided by 3.( i8 c2 w; i" U, l ?8 |
- | D# F/ K% z& @% L3 C* w
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 \- N7 _4 K9 ^9 [$ b4 Vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem; o6 ?7 [4 R% z$ a* [4 _
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)1 d( {. v) D- J
= K^3 + 3K^2 + 2K
# i4 d% L' o" J) q; u2 X = ( K^3 – K) + ( 3K^2 + 3K)3 x; f2 @( b/ p6 Q6 ~4 C: \
= ( K^3 – K) + 3 ( K^2 + K)
# ^# t! }% K; @! f: @) oby 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>05 T# T! o/ u* v1 P/ y+ W* U" ~3 o
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
7 h. X ^* d# @- m, U F = 3X + 3 ( K^2 + K)/ }* h0 O1 L" {5 Z' Q \7 X
= 3(X+ K^2 + K) which can be divided by 3
1 R+ }* U- _$ W
: r; D' j7 U" LConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.* P5 N( {0 ^$ ^& M1 Y0 i5 R9 ]
) v% D1 L5 N5 p& P% x[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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