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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)( ?& [, }. |3 b# `1 G$ C4 N9 f
1 t" {7 s X7 g
Proof:
9 Z3 }# B+ \- w0 F+ {0 e. z1 C( XLet n >1 be an integer z1 }0 Q7 s. ]9 J2 H1 `1 p
Basis: (n=2)3 @" d7 x* N# z3 k( P( I0 d' J7 ~7 }
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' P1 p- H! {* m/ q9 R( {. B
9 b& M% T1 q1 a. UInduction Hypothesis: Let K >=2 be integers, support that
" t t* c; h2 ^! s" N+ A+ R% v6 n K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
D! t ~ Z1 c' R* R7 ~since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem' w }9 S7 M8 i# h. J# ^
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
+ G8 s6 r) P3 U7 o, q" ?8 K0 o: @ = K^3 + 3K^2 + 2K
: R. ]! D8 R; V% M9 \( w = ( K^3 – K) + ( 3K^2 + 3K)
3 x K+ N: d% A8 _% l; _ = ( K^3 – K) + 3 ( K^2 + K)
# v6 ? Y# w! L* I2 |! W0 [3 Gby 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
% i% z$ X( q3 @# q. _1 ^1 B$ qSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)8 {& `/ B, Y& D5 X3 N8 a3 i
= 3X + 3 ( K^2 + K), Z R' ^4 f1 Y. M% |8 y
= 3(X+ K^2 + K) which can be divided by 32 C; B! d) a# l' |/ |) ? t
- m1 F" C2 V" D5 M1 L; M; yConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.7 ^" q% @2 q1 u0 u
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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