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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)5 h y! D2 U! e$ o4 R
H! k: m$ B) |. |1 H. YProof:
6 o4 m4 f3 i8 Q9 B% LLet n >1 be an integer
) X; o' P" J. H/ d. A+ r0 NBasis: (n=2)6 ^1 `' Z4 }. B5 L
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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* ]3 \1 {7 F% nInduction Hypothesis: Let K >=2 be integers, support that. a& F* V$ T2 @7 f; }
K^3 – K can by divided by 3.) x/ y }1 Q/ o. u
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3# J% Z4 U$ a1 G" U9 ?* t
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
) w4 B2 n: p5 _+ H" c }( IThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' w; Z$ \/ d0 A
= K^3 + 3K^2 + 2K% M7 d! |4 l* A% ?. O( K# `
= ( K^3 – K) + ( 3K^2 + 3K)1 G) j$ D# _, u. R& G
= ( K^3 – K) + 3 ( K^2 + K)
6 M i6 Z; I/ u: Y- ]8 v/ Jby 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 F8 ^; W! X9 M
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)& a' s9 O0 b3 v7 g) i" m& q* k
= 3X + 3 ( K^2 + K)
+ O9 U" O; z9 B$ e+ ?$ z = 3(X+ K^2 + K) which can be divided by 3) @) K6 V& K3 G6 k9 c* \9 S0 w! Z
( W9 ?9 B# U5 o& {
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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1 U& O; G, @! W) r/ q[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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