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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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Proof: , P8 B' t& L- K/ {, D) Y
Let n >1 be an integer $ X. [6 h0 y3 l3 }9 z( d
Basis: (n=2)
5 d7 Z' w9 Y8 n 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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. T' R2 L+ E2 b7 MInduction Hypothesis: Let K >=2 be integers, support that1 b( x+ J0 U) Z
K^3 – K can by divided by 3.
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1 m/ [6 Q8 K* y5 ~2 H& XNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 36 D: P L7 f; N9 k
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem& H7 ?& A+ r: `. H. N! J" O
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1), W( v( }" g2 B* l( U1 u
= K^3 + 3K^2 + 2K0 q( v, K, S# r) B0 a6 ?6 H
= ( K^3 – K) + ( 3K^2 + 3K)1 _. l( R# H' X, m
= ( K^3 – K) + 3 ( K^2 + K)
) s. _" T p8 R- Eby 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>07 i6 j( j9 J( n/ z3 r, ?' l6 x! |
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)& V( m3 _ D& T4 t
= 3X + 3 ( K^2 + K)
" n+ T+ A8 F' r* H9 w4 Z = 3(X+ K^2 + K) which can be divided by 3
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.* P( b) X" t0 Q0 g9 X' d" ?* Q2 ~
( w) D4 C& F! |, }+ _+ N, h4 J[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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