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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)/ `+ W4 B1 Y, q3 Q o
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Proof:
8 Q. X9 V, n/ J* c1 _& e5 PLet n >1 be an integer # O3 E) s1 Q. V- u2 u# q
Basis: (n=2)9 V5 k! I, e* j8 l6 f- e
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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8 i# D2 O" c" ]3 vInduction Hypothesis: Let K >=2 be integers, support that4 N& I) b' B) c2 w# e \
K^3 – K can by divided by 3.
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2 j: x( C- H% p. r8 {- ANow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
& e% t* W" g: `since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem- G" y; h+ O- @6 ?% k- F2 C
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)& F' Z8 P k% `: Q0 t* a* [2 w
= K^3 + 3K^2 + 2K
. m5 b8 I' N' G0 n) @% q1 i9 |2 W = ( K^3 – K) + ( 3K^2 + 3K)
7 l8 \. e6 O5 N/ P = ( K^3 – K) + 3 ( K^2 + K)
0 C7 ?' O4 J2 z/ z* f! i: Iby 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
, ]0 }% n$ W1 fSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
! R* w. r L" O& y' p- l = 3X + 3 ( K^2 + K)2 H, {$ I: A1 d2 ~) `
= 3(X+ K^2 + K) which can be divided by 3" P q& r1 d4 ~2 f( [3 o
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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