 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)+ g1 g% T3 N3 m9 l
- o# G5 \- K. P; f7 [: v! X
Proof: % ~6 a; V2 _. _+ ~/ L$ ]
Let n >1 be an integer 6 m( |( l1 L) a4 A& I* h
Basis: (n=2)
( @ L& r8 Q8 c 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
# i+ [( [1 F* t: K% _3 e. ]$ S7 s( j+ Q/ n! }2 \3 F
Induction Hypothesis: Let K >=2 be integers, support that
/ g& c2 t/ R! R* O; T! ^2 w) I K^3 – K can by divided by 3.
& j4 X9 C( d* ]/ r/ Q5 ?
; x; L, j# B4 P2 u9 eNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3, E, w6 V- y I( V8 ?( w
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
# O/ m( h/ N& k+ BThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)6 K8 [; Q" q! x8 l1 _9 v
= K^3 + 3K^2 + 2K& U! A( e9 v+ T9 h0 n% ~% m
= ( K^3 – K) + ( 3K^2 + 3K)
" E3 F; P* C7 ~- Q, [- } = ( K^3 – K) + 3 ( K^2 + K)
' j% t' m! J+ Y+ u. v7 Lby 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: F2 e7 R$ b L. q
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)) T' E5 p; k! R; `& n4 {
= 3X + 3 ( K^2 + K)
9 B( ]9 V1 T+ `* K = 3(X+ K^2 + K) which can be divided by 3
4 p1 p, i9 y8 V- [" R9 R4 d9 \
# q8 B( Z) Y& g) \, k5 ~6 jConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
0 b% R, O( N6 c4 h4 n4 h9 \ X C7 _6 g8 y0 f: G0 W
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|