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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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+ |. m: _$ c6 K; p8 d1 QProof: " E7 \. Z) ]+ i" j' J1 k- M
Let n >1 be an integer
' ]8 z! P; e \3 L0 k* pBasis: (n=2)
c! K# `& W* `" K 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# ^6 ^: f) V5 J
; n& ~! R6 m& [9 F8 ]Induction Hypothesis: Let K >=2 be integers, support that
! ^6 j: O+ x+ p7 T# U; U K^3 – K can by divided by 3.% a- ^7 f |% g, g$ X; @
* a8 ^$ |, x, p8 \( ]7 _4 T* k
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
- h$ x4 @* p; Vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem7 D0 {% n( d) v
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ A" o' |/ v* I9 {5 ~ = K^3 + 3K^2 + 2K q- s" F( Q% i, y8 n
= ( K^3 – K) + ( 3K^2 + 3K)
& k" \- i+ F3 c& { = ( K^3 – K) + 3 ( K^2 + K)
, `/ C' Y; ?3 M8 D# d0 k9 Rby 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
]" |* ?5 _, j( ZSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
3 S& u/ ?1 c% v0 x3 C& [) w = 3X + 3 ( K^2 + K)
: }' S! a$ l( C7 Y = 3(X+ K^2 + K) which can be divided by 3
& G; o; Z) M6 w2 W5 [- B; F/ _8 ]! l5 i \
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.6 X4 Z3 v- T; L8 e. ^
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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