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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)
4 ^0 j+ F% i* B1 {: g0 u- L/ b
b' Y1 q- L) U6 eProof:
8 I4 ^$ t% r" M& N6 dLet n >1 be an integer
3 q/ n, ?; Z9 t% W. I2 FBasis: (n=2)
4 Q3 C" L8 Y p K. }4 ^ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 37 |: c1 Q4 C6 T
1 @7 M6 @" o- J: X5 k4 zInduction Hypothesis: Let K >=2 be integers, support that
* e5 [) b0 h( t( ?9 I8 Y K^3 – K can by divided by 3.* d) q5 M! r. h& c6 c
- J& s' {1 Y% T/ G* RNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 30 B& ~. f# \# B
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
% H' n) C/ P1 M7 }, y5 @Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)2 V+ v) W4 d9 u% R
= K^3 + 3K^2 + 2K7 T( l8 r, K" x h
= ( K^3 – K) + ( 3K^2 + 3K)
" A" w: c, Z# i$ e, Y = ( K^3 – K) + 3 ( K^2 + K)
" H& p- F7 {2 z8 rby 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
0 e9 D- a: B2 B3 O% I2 g: q( C' uSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)5 R5 h" z' N" F
= 3X + 3 ( K^2 + K)) C; J# m$ S" e5 g
= 3(X+ K^2 + K) which can be divided by 3
! X! ?! h- [/ n6 ~0 @# x8 U. d8 D7 U# D2 n9 i C. I
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.1 G. e1 Q9 C' H
$ K) _9 w( Y- I8 E7 v) F[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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