 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)! d5 y" M7 {: ?7 |6 T8 X- s# d
2 |. R% m8 c) r* BProof: $ j) Y* W& H L
Let n >1 be an integer 9 @0 y! M9 q7 J0 ^" I6 f
Basis: (n=2)4 ]- Z6 p* Q1 N) |
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 34 T! X) q& R' Z/ T0 T$ {4 a
# |1 ]3 M, n! t h
Induction Hypothesis: Let K >=2 be integers, support that
/ l! p3 H8 e; {" c7 i2 N K^3 – K can by divided by 3.1 n, Q! v3 V9 X5 V" B1 N- T
* x" L- ]! {2 X8 F. g% Y! ?
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 30 Y y& X9 ]! s3 ^' O
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem! C7 P+ E# e+ D8 z# J7 U6 [
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' w0 L5 K$ G3 W" Q
= K^3 + 3K^2 + 2K% b1 {" _' b* C `% n8 J; L
= ( K^3 – K) + ( 3K^2 + 3K)1 R: G: u: ]* u9 g6 t! y/ [( ?
= ( K^3 – K) + 3 ( K^2 + K)+ r+ i, S' Q4 H- 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>0
! ?/ ?; d$ |# h$ s# WSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
- u% ~0 Q1 V) U3 Z0 S = 3X + 3 ( K^2 + K)6 u& ^8 x& \$ h6 }# Z* A# e+ ~
= 3(X+ K^2 + K) which can be divided by 3- T+ F2 [; x! e; E% y3 `& H
) W# v$ _$ P3 [+ F/ B3 r% D; \8 a
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; E. L8 F% i5 C5 V' F( a6 n7 V( y
+ s; W/ t ?; ~) N) m9 s$ v
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|