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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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3 Q7 P7 t5 _" z$ w0 RProof: ; u+ E2 ?4 \8 b, M7 Y& B/ q
Let n >1 be an integer
" q* \. m1 i( |5 c& h' eBasis: (n=2)- x) R. N! C% A$ {! u+ l4 O" U
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/ t0 ^7 H* ^$ r2 p/ E0 Y4 G
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 32 u. x1 D, z3 {' b3 u+ W, H- v
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem% S A0 h# K$ v) @3 U' {% z2 x1 c
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1). y% Y4 }) @: H4 W0 j
= K^3 + 3K^2 + 2K
# w' |6 V+ F8 u9 \2 [; s+ n" @ = ( K^3 – K) + ( 3K^2 + 3K)1 d4 b& q9 D( u
= ( K^3 – K) + 3 ( K^2 + K)
% M! j% f; v) Q1 rby 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
% x& _5 J( N6 j& U$ A- r2 ^So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)$ l* n5 B+ T- E% i/ U
= 3X + 3 ( K^2 + K)/ X: c3 ]6 m; g3 M7 K5 { [
= 3(X+ K^2 + K) which can be divided by 3. J# M( D7 {" e4 t+ H. n
3 t% _! M( G" ]# l, L% x9 V: WConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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8 A% m2 Y6 S+ B1 X2 W1 x$ ?[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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