 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)" O% B5 d" ~! |6 x) O6 |' l' E. `
" w$ K, i- n& O" e. d7 h: g+ c
Proof:
# r. U2 l6 s, TLet n >1 be an integer 6 @& Q0 L: y- D4 l
Basis: (n=2)
# C1 ]" E! C% g 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
) x# R* L8 h. R
6 D" Q: y7 Z& v4 N9 b# e, RInduction Hypothesis: Let K >=2 be integers, support that4 N3 @5 h4 m# Y" i5 h
K^3 – K can by divided by 3.
/ I2 C( L3 c7 ^2 t( t' p9 k/ [) B2 q' L" Y s: D7 `, N7 v
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3( \4 C* a. \. E4 ?
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem* Y& T) c9 v7 M$ J
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
4 e6 M& _8 s7 [" E+ A7 S" W1 g = K^3 + 3K^2 + 2K: m3 O& [, ^) Z: R1 F/ \
= ( K^3 – K) + ( 3K^2 + 3K)3 ?0 p" ~+ F0 m$ m/ g: n
= ( K^3 – K) + 3 ( K^2 + K)
6 X0 J$ I. [' w' V! x' Vby 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, `9 z2 [* v& C* I; s0 [
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)- c4 \$ C: a. o% P T l L( J! y
= 3X + 3 ( K^2 + K)
0 @, M: l% e/ h3 }/ ]* X8 y = 3(X+ K^2 + K) which can be divided by 3: `- n9 N2 y. a: B
" o7 ~) g+ l' f3 g2 u
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
* Q9 x6 j/ ?7 H8 `3 P5 n" \$ c4 [; p
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|