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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)# j9 ~- i) ?. ~( H* Z
) @, Q( z4 c. k# J
Proof: ' h. r9 H- ]1 [' H2 J$ @
Let n >1 be an integer
; n7 s8 H( ^0 L7 GBasis: (n=2)
& a# f4 s/ a9 E. { 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3& f7 z) A7 `& r* K
' [+ `, d. \* K6 Q1 z' TInduction Hypothesis: Let K >=2 be integers, support that) f J; R0 g0 h. q
K^3 – K can by divided by 3.
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' @' T! W4 c( Z. sNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 33 x# A7 i# W1 `7 V/ w
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
5 R% m) L1 I3 D/ cThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
( [' b, r: \/ Z4 c/ p- z' G = K^3 + 3K^2 + 2K( r W3 {2 r5 x: q) A& [
= ( K^3 – K) + ( 3K^2 + 3K). R2 @% l% B$ B( L8 t5 M( z
= ( K^3 – K) + 3 ( K^2 + K)
0 |: @' S2 I( O, q1 U; a, Pby 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% H c; i N+ L, Z
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
0 @' J' F8 ]; _( a' } d = 3X + 3 ( K^2 + K)3 Z, F+ W, B+ d2 m, g
= 3(X+ K^2 + K) which can be divided by 3& }" U; ?2 m. s' j+ m
% u1 }3 P! B7 m/ G+ j: K) I0 b
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.2 k. v. F% u& [, \" v, v
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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