 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)9 Z0 _9 F; M# |: _( m
5 C: j, W4 V2 V# ~) KProof: 8 L/ C0 F5 m& S
Let n >1 be an integer
% m9 U4 U5 z6 h6 `+ d6 @Basis: (n=2)
& B- `5 g2 y6 P! _7 B8 C 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
8 p. z! b3 K- e$ S8 \8 j' B; b# R+ ~8 i& D& B% X% \0 v: l
Induction Hypothesis: Let K >=2 be integers, support that0 }/ S, d: G/ k/ x
K^3 – K can by divided by 3.7 p3 \& k0 g9 E' J
# R; U6 a: x9 U8 r
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3& l' n+ U4 y- u. m/ o2 Q) m3 m0 R" _
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem7 z9 i. o& _: r! ?1 D2 b2 U. N
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
5 }; \1 C$ c/ j/ `- f = K^3 + 3K^2 + 2K
6 B9 l$ r, [2 Z, r = ( K^3 – K) + ( 3K^2 + 3K)% F& j$ p+ t$ H4 Z/ { V
= ( K^3 – K) + 3 ( K^2 + K), q* V. V) Q4 P8 \9 d# s
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
" t( n; |$ L+ [5 l- ]. M$ v8 {- WSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
1 l' i: G4 G8 F = 3X + 3 ( K^2 + K)) S$ {/ ?9 H: K: q' m" N
= 3(X+ K^2 + K) which can be divided by 3+ X# _8 B+ \) l
q. j/ }3 A8 }* U( PConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
! t. g" i# c& }- f5 B
" g4 _1 D8 h% x) \& o[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|