 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)6 ~* l) c9 c( q \
7 s9 J: J& M: I) K/ `( N1 ^
Proof: 3 Q: o! {$ d1 [) x
Let n >1 be an integer 2 [; E: g9 C! V- B! _: T9 ]# z
Basis: (n=2)4 c" y8 i, X( k7 K
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
2 Z8 [8 K8 D5 U$ n# K+ F' k$ W. H/ z' |2 b
Induction Hypothesis: Let K >=2 be integers, support that' A$ T7 L1 L" S m6 L) X
K^3 – K can by divided by 3.
# C6 ]# a. Q4 c4 v7 \, N# i/ u, H, }3 r- e( K
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, w( n# s" \+ x7 Vsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
& X7 w; n: `: `, ?Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
2 p$ m z, L7 H T8 D8 B/ |4 R$ ^ = K^3 + 3K^2 + 2K
/ X7 C' g" i, l$ H9 n" n& ? = ( K^3 – K) + ( 3K^2 + 3K)1 p6 E6 {7 Y, Q- I
= ( K^3 – K) + 3 ( K^2 + K) E8 |- b' O- b/ |
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>04 S4 Z3 y2 a/ R3 j: `- v- D
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)4 P* m- D, H: E# Q: V& t
= 3X + 3 ( K^2 + K)0 X \% o0 h5 B- e
= 3(X+ K^2 + K) which can be divided by 3
. V0 |; |. w1 g2 C) i. N6 I; X! Y$ `+ J7 {, q) f3 i( l$ m# H) G
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
, E' R/ s I. ?( e. |( N. f9 [$ f6 W! S. C8 t; f+ [' T4 k
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|