 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)
* g9 ?6 D; D, S1 M- C
- w* E" u- a) H- [! C1 d9 gProof:
- ]; z0 g: q9 Q( g/ z! ~& v* e& g% iLet n >1 be an integer ) o/ J% P2 j/ s# @
Basis: (n=2) j2 x; L4 D# W
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' _8 m2 w" n& r4 K5 b: \0 `
! D% W0 }# v- @Induction Hypothesis: Let K >=2 be integers, support that
& _, ~1 O8 E6 \' X K^3 – K can by divided by 3.
j7 ~' a2 ?9 Z8 k7 J8 M
" U$ Q0 J$ H7 ]( }: f& |; HNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
' H8 O! B* ?; O5 }since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem8 r [* F& K7 E. U. B( i
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1): a7 n0 |) E: I, D0 Z' n
= K^3 + 3K^2 + 2K' @7 }& O5 D7 D1 q2 B2 Z
= ( K^3 – K) + ( 3K^2 + 3K)
# r0 @- T# A- i+ f# v3 Q = ( K^3 – K) + 3 ( K^2 + K)
% l1 S6 n: K& |) h) |% v+ d! }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& w- y) x' R# [. j. Y0 ?
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) I L3 v8 k% ?. v = 3X + 3 ( K^2 + K)
2 W! f6 s1 w0 T, h = 3(X+ K^2 + K) which can be divided by 3) B! R: x/ i E8 j* S' `1 X
% W8 S3 o: P3 b* W! ?Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
' v. T N1 _5 {7 \, _8 U+ c* a/ [, u2 i5 _8 Z, O: ?4 ]
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|