 鲜花( 53)  鸡蛋( 0)
|
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)5 P3 J2 I7 ~. c% ]2 A) f `" l
4 m# {+ i" {# l5 bProof: 9 N" z8 q* ~0 ~9 {5 z
Let n >1 be an integer
5 W2 B/ _- O R+ z0 z% y% s8 pBasis: (n=2)# z$ _% O# F0 G2 x& i
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
4 Y& H4 ~6 [1 e- i3 n3 O/ ]2 t& H
2 v$ M4 ^: Z4 f# I) MInduction Hypothesis: Let K >=2 be integers, support that
: f7 c/ |, l, a! S/ Y K^3 – K can by divided by 3.( M6 X' b5 v9 x5 _/ D( P0 {9 F
& P% c) v( k1 p: U( v1 `# o* D) LNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ r/ I- l, G8 W @# m- b* I4 Fsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem1 g9 S a3 Y' W( P/ ~; _1 M& [
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
2 M# s# p! O* F: D = K^3 + 3K^2 + 2K
, E! m# M* F" t6 B = ( K^3 – K) + ( 3K^2 + 3K)
4 |" ]5 j! ]/ i' g = ( K^3 – K) + 3 ( K^2 + K)& q$ e7 r/ U4 X" Z" n2 @
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>09 s9 h7 R# Q7 c, @. Z8 ]( F! I" h
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
# e6 B" ]/ {0 S6 y# O5 Z) ~ = 3X + 3 ( K^2 + K)
/ d8 z$ p Z9 u0 @/ x% X = 3(X+ K^2 + K) which can be divided by 3. S8 s% s& D( i6 F# _1 [
: ~- R7 {; ^8 K! K" DConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
5 z% O4 l2 p8 e3 u9 }9 k7 h: D1 @2 z7 O8 t- |, `7 v7 u
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|