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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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$ G+ S0 r. R% Y1 n+ Y$ p' T+ O! ^Proof:
" Z$ s& l0 s4 Z8 S0 ^% WLet n >1 be an integer
; T4 \7 F) s6 s7 t: {Basis: (n=2)6 u* P3 {. B9 b8 Q0 a8 @
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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- }! r& O& N( F2 k6 xInduction Hypothesis: Let K >=2 be integers, support that3 S) n9 z H0 {
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
: [5 m4 }5 n( m v5 A0 E8 I& J- gsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
D2 {" N+ i# ]) V2 oThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. E: ?- a" j5 j- u) Q! j" K = K^3 + 3K^2 + 2K
1 p# Z. t5 q- B/ M( } = ( K^3 – K) + ( 3K^2 + 3K)
5 |1 b. U/ Y% [& [0 y, a+ T/ { = ( K^3 – K) + 3 ( K^2 + K)# o# ]; I* s* z, H/ p* b# ^
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$ P- U$ A! z$ J" z) p9 l
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
8 C$ F) C; ~# V, v, U = 3X + 3 ( K^2 + K)& n2 ~$ w7 h1 W% t3 @
= 3(X+ K^2 + K) which can be divided by 3( Q3 v3 t: O# V5 ]5 C2 P
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.% Z6 o1 c5 n# E# z8 z( ^
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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