 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)- [3 ~; Y7 q# C& h
. h+ S# e% C6 O, h. A
Proof: ' v( W7 }$ ~9 X( l6 i& U7 c) h$ ~3 C
Let n >1 be an integer 2 P/ g. S+ L" w
Basis: (n=2)
2 q8 t5 Y# S' j3 N. y% M( Y 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3* J6 X7 P( a0 B+ _. h3 O
8 b% q! b- N$ q) g
Induction Hypothesis: Let K >=2 be integers, support that" [7 n2 o* |+ P, w, n/ x" i- d
K^3 – K can by divided by 3.! Y" l: p: s+ x; w* P* K) F5 `
: b# i; v6 I Y& b3 l1 X& O5 I9 ONow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
K' L5 V1 l2 @: dsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, ?4 I" I1 o4 [* P! kThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)) }3 C% i' |. ?# ~$ }
= K^3 + 3K^2 + 2K/ _% G/ s$ w* k
= ( K^3 – K) + ( 3K^2 + 3K)$ n% f. g" R9 s1 y& ^
= ( K^3 – K) + 3 ( K^2 + K)
; p) g* B6 Z% R) m' ?: {3 Jby 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
$ p8 r% z3 M- I+ `, lSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)# H! i7 e9 ], t8 X
= 3X + 3 ( K^2 + K)
3 Q8 x$ @( `( W% e5 J. m! @ = 3(X+ K^2 + K) which can be divided by 3
9 A$ _) G8 @9 f |, O# T; t: ?( M5 V3 [
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
3 V6 T) Q _, X
. n% S$ t2 S% M5 W# o9 C1 R( }[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|