 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)
. Z0 {; B7 g! }3 `, e5 I( P) T
7 A. B0 t: W% ]5 `" {- ^( z' zProof:
0 a5 y% \. N+ Y5 d, ^Let n >1 be an integer
$ [: J! b$ A& t1 OBasis: (n=2)9 R2 i3 |6 S: I6 Q y
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3/ V5 r( y) N2 z
% J8 y4 N8 O+ i$ b6 p8 FInduction Hypothesis: Let K >=2 be integers, support that2 ^6 g2 d; U' H3 u/ N2 c7 I# O8 |, _
K^3 – K can by divided by 3.
3 U* W& R, x2 X* b5 n
' D' h0 z* R6 k% f4 v8 E6 m2 }Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 33 `8 S$ w: C) t
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
9 K: Y0 s- r3 |5 V5 {Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)! L3 E5 Z/ D' I! P
= K^3 + 3K^2 + 2K
5 Q0 O, a- J% i8 w+ b3 I+ O) x = ( K^3 – K) + ( 3K^2 + 3K)0 v% U$ J4 d( R- u! H: a- f) I
= ( K^3 – K) + 3 ( K^2 + K)- f5 l% U" n2 [ h& V
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>0
5 y7 u7 g( _5 K* W6 d+ ~" P2 @So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
, @" }* H; P, r8 x7 s" ~) ^ = 3X + 3 ( K^2 + K)( \1 Y- F( I4 p/ Z, y9 S: Q6 m
= 3(X+ K^2 + K) which can be divided by 3
6 S& b2 D! T v& u3 Y0 O0 @+ a5 f* M* W( ]5 d- {' \5 r. H
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
+ O9 |2 u) S: z" d
+ v7 A$ e9 p7 d1 P7 M$ n0 a8 S[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|