 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)
/ v2 G5 H# {4 \/ S
9 `; N/ Z8 P, o$ a2 L- bProof:
1 _0 d7 i- O5 WLet n >1 be an integer 5 i7 w- w. x4 w- ]6 Y
Basis: (n=2)' j& o+ b2 Q/ w4 o; C
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3( E8 I4 L6 {* m' O7 D! [7 K3 N( K
) y" n+ l" D8 e& c( P- hInduction Hypothesis: Let K >=2 be integers, support that% y: O2 N& T' _+ ]/ ~. ^
K^3 – K can by divided by 3.
* D: G2 ?2 ?) C5 _5 E# A7 {, C; ]1 y: \6 p) y/ ]# }
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
" {% D# \! G& I4 C, o; \$ W" Bsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem7 ?4 g% e: c9 _, N0 G8 C- T$ g' G
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)( m! w: {7 X: g8 C# r% z* B: a
= K^3 + 3K^2 + 2K6 L2 O+ x' @: p# i
= ( K^3 – K) + ( 3K^2 + 3K): k6 o$ {; {& k; B4 X# ~* V1 K
= ( K^3 – K) + 3 ( K^2 + K)
( N% E' @- P$ Zby 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
( ? R/ |3 U2 g; W# q. SSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
' |. F& m, A4 v3 J+ d2 b1 p& ~7 ? = 3X + 3 ( K^2 + K)8 s& r" S. ?- g$ o) L p
= 3(X+ K^2 + K) which can be divided by 3' n* [0 [2 B- N2 b6 n. G0 Q/ ~
8 j M" @/ N4 X9 l7 M! M% x
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
4 \2 t) H& H0 w9 i0 ?4 T& _- X& U0 D7 Z ^
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|