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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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9 g- u, Z9 u* n5 x6 qProof: & x. ~) |5 P n& ^
Let n >1 be an integer
5 i" Q& |) _: Z% Z( m' ~Basis: (n=2)
8 T: G4 W5 z! y8 Q5 z 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3 }2 s0 r5 n. ~% ^
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Induction Hypothesis: Let K >=2 be integers, support that
& p9 W* R; {. F) S0 h' q! G# ]# { K^3 – K can by divided by 3.
: A. \$ K! \* c$ l6 W1 R; Q3 X W7 M7 N) i+ k/ {+ ]
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
4 `3 F( W* B6 }since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
$ ]( h# J! G0 ^9 ~& k# B9 jThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)" T8 D' q" O* Q" M: j. R# |
= K^3 + 3K^2 + 2K `* G' D8 c! ]3 e, H/ c: O! k5 Q
= ( K^3 – K) + ( 3K^2 + 3K)7 j8 Q* O% c7 }8 \7 ]; k9 w
= ( K^3 – K) + 3 ( K^2 + K)
6 G3 _' A/ _6 X) Gby 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, r4 [$ m. O9 V. N; Q, f9 W
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
2 D* @+ v8 o' C = 3X + 3 ( K^2 + K)
9 U% o( ?- U: J = 3(X+ K^2 + K) which can be divided by 31 v8 U% g1 C0 q. Z8 F3 K3 K
- ~6 E3 ?2 h# Q3 Z2 aConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.! C; |1 y7 Q, r) H
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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