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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)# P7 V9 _! z2 V: B
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Proof: + V. v- o s, K3 ?. ~, o
Let n >1 be an integer . K# ~1 Q7 M; G( r
Basis: (n=2)
, H9 m0 j" I6 a, S2 B 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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- U4 l$ r3 Z) C" F% k0 B5 zInduction Hypothesis: Let K >=2 be integers, support that5 X& G$ H. }1 h4 L$ v
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 3
+ Z/ I) q( d7 L3 J! V; ^since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem" v( c6 x, o6 [9 n! E |7 w
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
; ]- B( i& E& O8 r. ^ = K^3 + 3K^2 + 2K2 _1 V4 ]( w8 W& I" u3 a& ]
= ( K^3 – K) + ( 3K^2 + 3K)/ x: Z" N: d2 h1 m
= ( K^3 – K) + 3 ( K^2 + K)5 r: K* X& V- E8 x2 R9 W% H2 F
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>05 \" T5 _1 S. S6 V+ r
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
( \, o6 l% P4 M1 q; Q% { = 3X + 3 ( K^2 + K). o. t+ p8 {* n+ l) }
= 3(X+ K^2 + K) which can be divided by 3( l8 L. V, V7 K. h' D
# l: o w+ Z* yConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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4 { s# i& d2 d! ?, ?; Y3 b0 E[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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