 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)
' v$ H5 v+ ]" K" X k; q. z4 x$ h$ J0 S5 g0 L4 C+ h
Proof:
/ D e2 b/ N+ |8 ^* H0 }& u2 FLet n >1 be an integer
9 M! P6 U7 o/ d& p! s9 S% yBasis: (n=2)5 T9 a# y9 D! M, V, e9 q
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
3 t; m7 N/ `8 t' [/ r" K7 v0 b1 X
2 l' k3 n' P, t. n+ g2 |! o' FInduction Hypothesis: Let K >=2 be integers, support that6 K2 Q; F8 L0 z2 }
K^3 – K can by divided by 3.& t8 Y% G/ g" Z# k
6 b5 l) b' N9 \# @$ o+ UNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3; y+ W: u0 P9 c' m& F
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
# Y W+ D! N7 ^" C/ s& _Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
p6 k" W5 E4 b; \5 f D# u = K^3 + 3K^2 + 2K/ X; p* v& G$ D+ p- H1 w
= ( K^3 – K) + ( 3K^2 + 3K)
+ V$ r0 s- k, k4 R = ( K^3 – K) + 3 ( K^2 + K)+ W' d& h+ s( L3 v2 u9 D% ]# q
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& @2 p+ i6 o% l$ G, i
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)# b! w2 b4 F* {3 m) w+ m
= 3X + 3 ( K^2 + K)
& D% \) I- ?' U8 F6 c! F% W6 u = 3(X+ K^2 + K) which can be divided by 32 _( ~, H) K- e$ g4 H7 l
( V4 B; p- t% vConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
2 R2 ~& ?+ J4 {! b3 c- I* W8 h
5 m8 p' R9 q5 V7 w2 {1 q[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|