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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:
& o. ]5 G# o0 v0 }( S6 OLet n >1 be an integer
% [ @; Z3 H2 ^0 ^" U2 v7 v% i9 N" vBasis: (n=2); f7 n% A( c4 y0 i
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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* t: ]0 Y& b" b' ~# G6 _Induction Hypothesis: Let K >=2 be integers, support that
$ x) y0 A1 t4 \/ m$ U8 F7 ? K^3 – K can by divided by 3.' Z7 b% z1 S) l; w8 d3 v0 ~
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
( B* D( |) C F* `/ a, isince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
6 Z# K# o# ~* Y) ^1 t; Z; u5 ?Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
" c& Y* s( V2 ]: ?; w2 ?% r = K^3 + 3K^2 + 2K
% Z7 ?8 c# ~8 D# L5 _+ O o$ q = ( K^3 – K) + ( 3K^2 + 3K)* o$ ~0 J) R T- y
= ( K^3 – K) + 3 ( K^2 + K)
& N" l0 K6 J) c1 d) a# \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>02 Q _5 h7 D6 I1 J* Q) l8 J
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
0 S& R6 N, S6 k1 }+ D5 N = 3X + 3 ( K^2 + K)
, y0 X+ V6 t6 M8 Z+ V5 _ = 3(X+ K^2 + K) which can be divided by 3
) ~3 D# P9 y/ G+ h0 L# T# b+ y) n
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; }. G! U1 N: L+ W
, K7 ^ L8 I4 w[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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