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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 t. ?3 f' d, s5 P' KProof:
( J0 I* k/ E6 t9 VLet n >1 be an integer
, C' N8 N- m1 Z3 j# vBasis: (n=2)
2 i5 X6 f7 Z* S4 G 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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9 R( z2 @/ \% C P v* NInduction Hypothesis: Let K >=2 be integers, support that
8 [1 c! I) n/ [ K^3 – K can by divided by 3.& J1 V. N4 {0 C0 E: ]) ~8 z
. |6 o, m1 X! B9 s8 f# KNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
: z; ^& A# n2 ?since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem% P7 K/ d' U/ k! y% l) C8 a
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1), d, ?. @) N* Z" Q8 m8 T K( s
= K^3 + 3K^2 + 2K1 X l, @* \$ n- R6 t7 a" E% d
= ( K^3 – K) + ( 3K^2 + 3K)
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+ @1 G0 r3 D; \/ \4 z/ q) Sby 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
. R8 B1 r) o. B, v0 KSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)3 S7 Z; n8 v! o& d z6 Q1 k8 A
= 3X + 3 ( K^2 + K)0 S2 O, h" ?3 ]$ I
= 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 >1.8 t; e' O3 G+ P! E; r
7 E* X y! }; ?8 y& W[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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