 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)) b4 t$ R2 E% @% o, \: Z
, `" G! `6 _7 Q, z5 ]Proof:
0 |6 }- \5 I( r8 w) u) Q, n- ]! DLet n >1 be an integer * O0 \) P5 f3 }
Basis: (n=2)- h$ g. q1 H$ W
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
! S P' }4 H3 r4 h# v$ e6 S' X! F$ T5 l: p. T- A* b7 Q! q
Induction Hypothesis: Let K >=2 be integers, support that2 f0 X2 u! k1 [" n$ v, A
K^3 – K can by divided by 3.
1 L0 U, E& r$ w# ] @$ ]$ Q- j2 ^6 a9 G2 U
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
9 Z: ?9 o) A! k% e3 T; Q2 J( }since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, r: c( A8 z4 Z( A
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
' e: o# f) @3 j0 O = K^3 + 3K^2 + 2K
6 z ? ]4 S, |5 F8 s. P- C9 J = ( K^3 – K) + ( 3K^2 + 3K)
' f- E; o0 [$ {+ w& q: G = ( K^3 – K) + 3 ( K^2 + K)
2 |9 t' Z$ ]3 I" [9 q; Fby 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
: F3 q0 T& T" `) n! U8 q% r" iSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" h& f9 M, y6 r% S = 3X + 3 ( K^2 + K)$ D# l. |1 k' i' {; k& t5 A8 ]2 M; M
= 3(X+ K^2 + K) which can be divided by 3
( V, {% j) M% t6 ?
$ c4 ^, v4 m4 n/ z7 dConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
* ]1 ]% L- C: ~" r8 M3 k+ j
! D" D$ e% W5 b[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|