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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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6 v8 @4 d z v% g3 j! d6 g" B9 RProof: 6 l7 L8 s; D' H/ A5 e+ A1 P* P
Let n >1 be an integer
( \/ K- L. R) ?Basis: (n=2)
$ {' l7 ?9 H# a1 |; j 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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6 w8 g7 z- Q: m5 ^Induction Hypothesis: Let K >=2 be integers, support that8 h: y( E: V, v D
K^3 – K can by divided by 3.
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3 b+ A; {: e/ c6 KNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3' t0 w6 J* o$ Q1 A. ]$ t& {
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem) ^+ @4 \" A7 i3 U
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
1 o' F/ p, `$ o* x) H! U = K^3 + 3K^2 + 2K* m8 `# h6 [7 t8 T" H+ Q: `5 W
= ( K^3 – K) + ( 3K^2 + 3K)
# }+ F! {( D& r- }# h1 t2 m* L = ( K^3 – K) + 3 ( K^2 + K)
# J0 m4 j9 a; h5 [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
6 X, _: V1 C" W% T6 I2 ISo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ y) Q$ a8 x* V" e9 i% l = 3X + 3 ( K^2 + K)% A9 A/ A( P5 W e" n
= 3(X+ K^2 + K) which can be divided by 3/ f+ l5 Q# E( ?% B% m
( U* }* I( k2 ~' T8 v7 w2 }) zConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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, }( ?8 H* A7 b+ V6 f[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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