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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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5 z) P* G/ J+ s' f* F, B& Q$ NProof: , G7 R" F1 i6 l
Let n >1 be an integer
' W+ v, @5 K. N8 Q% o3 SBasis: (n=2)0 T: q- r# I' I! J {7 v6 ~
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 that5 L5 h& b4 h- D% P+ Q
K^3 – K can by divided by 3.5 k1 G5 B' g2 _* V
8 ~( T$ @2 K+ E+ z8 M( m! INow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
* f" |, `3 o( J4 Csince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem4 Z- L7 r% a3 _
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)5 m: P/ V. s$ v% v/ W1 [
= K^3 + 3K^2 + 2K
. L$ M+ ^ N X. H ` = ( K^3 – K) + ( 3K^2 + 3K)
. y/ m+ S6 d) a& H w = ( K^3 – K) + 3 ( K^2 + K)
J0 P7 W: A& N( y- B6 k5 Y* cby 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* q, H9 O: |3 C' H# f# \
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)# N; ]" Y5 I7 ^- w- a4 M" a/ R' g
= 3X + 3 ( K^2 + K)5 d( m+ X( d7 A. A. ~
= 3(X+ K^2 + K) which can be divided by 38 x4 G7 p9 W7 x/ y! m
, k5 I; \# B) J3 H. q3 QConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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5 g$ Q1 f; ?7 }, v' B[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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