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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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L: e/ V6 I8 d5 W9 a" U' q- fProof: ) X( Q( ?; u, a0 F/ j* {
Let n >1 be an integer
1 g0 U! V% c3 t2 Q4 DBasis: (n=2)
# j, T1 m& n% w. I0 K 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3, _6 l2 [. A+ z8 L# F& Z4 Y
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Induction Hypothesis: Let K >=2 be integers, support that
8 f# P% o, [ r K^3 – K can by divided by 3.$ S& d: a3 Z0 \9 r) j, M1 F' G/ k4 y
+ L) h# a! K7 @$ F% ^
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ w* O9 Y u( s0 Z* s" p+ u0 Q/ \since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem3 r5 |. L1 t/ y2 Y
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)( h. ?" X% u7 U2 A2 F$ a# h8 B
= K^3 + 3K^2 + 2K
* X; V+ C1 Q# S" `: u = ( K^3 – K) + ( 3K^2 + 3K)* x! V6 h' l* |5 U# M, k/ X
= ( K^3 – K) + 3 ( K^2 + K)
* [: p+ w" ^+ ^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>00 h3 U4 ]4 Y( d- T
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
; k8 J/ r7 G# J7 B" C; { = 3X + 3 ( K^2 + K)! Y+ o; ^# _* V- P2 P
= 3(X+ K^2 + K) which can be divided by 3" x* q& L- L! n, O3 w( i9 P
8 t; |& s3 j) o, }/ LConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1. j, n" p; z8 l/ n
/ B- b$ o9 O5 B. @# c[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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