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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)- A6 K# H Y I% E. L& E
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Proof: & k; w7 ^8 H, z. O( u
Let n >1 be an integer 4 F* H- x) P. U6 B7 ?) v L
Basis: (n=2)+ F1 n$ w) D2 F$ I
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
) R! j2 k! ^. W5 Q- @5 O4 R K^3 – K can by divided by 3.. f7 d1 n$ b6 {$ f: f+ a2 ^
9 [0 L) T+ X. @9 wNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
1 K2 ?' @( [( W3 u: psince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
& ?$ q" Z7 A; K' zThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1) u3 a+ _1 L! {
= K^3 + 3K^2 + 2K1 S4 L6 `1 R/ X; n) u
= ( K^3 – K) + ( 3K^2 + 3K). j- E/ h0 P7 Q& \5 @
= ( K^3 – K) + 3 ( K^2 + K)
+ |: g0 q; t2 Y) R3 W" q4 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
9 p5 Z- ]8 p1 ?. d( x4 ]0 JSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" K5 N4 U3 m+ f& J = 3X + 3 ( K^2 + K)+ e5 J! m! k- s
= 3(X+ K^2 + K) which can be divided by 30 k$ Y* S, ]' }; c$ V# V0 b; q M
/ J0 a! t# p+ \/ H- J1 M) x
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.- r& U- U; ^4 x. B; }% b- a, c8 O
/ m8 P b6 _5 H3 N; E6 O6 I% H* U[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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