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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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, Y* r4 ?7 V% h! O: V) c( x0 e) IProof: & J2 B! m; d* U) v+ R3 Q8 S
Let n >1 be an integer
9 m) h1 H+ d5 \* F$ ]! v# T1 ~Basis: (n=2). M1 }3 t" a- Q* c$ k" S
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 }) ^3 d( J$ e3 j& y# i
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Induction Hypothesis: Let K >=2 be integers, support that. N8 w6 h" \) ^+ ^! I7 u2 S
K^3 – K can by divided by 3.' R$ C% e' C" b$ {0 \9 \
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3/ L% {2 ^5 M3 ~2 J9 b
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem+ | ^/ C2 r/ z, |9 Z
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
! M5 Y% p. C/ m" |2 J = K^3 + 3K^2 + 2K; ^& Y. j# j. `# j$ e5 P: S
= ( K^3 – K) + ( 3K^2 + 3K)( v5 u+ W* J4 a2 y B7 C/ Z
= ( K^3 – K) + 3 ( K^2 + K)
, B; _% J3 I- O' e b2 Aby 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
3 U/ c: y3 Z) h2 D& m9 ~4 MSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)* e( U& k: K$ e8 z8 q
= 3X + 3 ( K^2 + K)
' a+ t9 R, ]# J+ C2 S# ` = 3(X+ K^2 + K) which can be divided by 3# p/ Q) P' S+ U4 y& b$ s9 y% Q
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.* ~4 R* K* @1 r& s8 g- O
. r9 G0 p3 n$ P: q& h[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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