 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)
# g, t' h# `) q0 L& `
3 e h* }7 J! xProof:
$ W- _4 W2 G# z: YLet n >1 be an integer
# X& b' x- w; j' V0 f1 o' Q+ FBasis: (n=2)
9 J4 C% c9 C, L 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
6 W8 J8 s" o, {* y
' D i L- t/ a. m% ^Induction Hypothesis: Let K >=2 be integers, support that: |0 O% ]1 F" p) E- E; M
K^3 – K can by divided by 3. F( z7 t2 t4 F0 f4 y* t
0 y" Q! `' d/ v7 gNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 31 B9 n4 R3 o; t2 u9 K
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
( f& y1 y* I3 L/ gThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
9 B" ~! c! K) u = K^3 + 3K^2 + 2K
$ [( h$ C+ O# n, T/ B = ( K^3 – K) + ( 3K^2 + 3K)
4 S' R" Y! s# a, H& o: G = ( K^3 – K) + 3 ( K^2 + K)8 X1 B2 X* F, B" d2 ]3 M
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 Y# m, r8 h* P8 J
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)8 s# X7 @0 L8 O+ a$ Y% d
= 3X + 3 ( K^2 + K)
: P8 X3 g+ j" Z1 [* [& E = 3(X+ K^2 + K) which can be divided by 34 |# K( }/ M: Z& j+ L) A& d9 d. D
, }, o2 O' s8 h
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.! `3 K. h+ H3 c8 U$ l9 U
0 ~3 C# P% m. P. Q' |' ?
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|