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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)
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) @; m G i* |9 CProof:
& E/ {+ p: M: |. W c) U1 aLet n >1 be an integer
- h0 R5 S. c5 f4 N6 @Basis: (n=2). K( P* u9 _: H- X+ J7 G6 ^
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 30 o5 z9 U4 ]3 ^: ~; Q" i' O. J
! q0 v! t I0 {* f0 ~Induction Hypothesis: Let K >=2 be integers, support that1 O5 k% k, f0 q8 }
K^3 – K can by divided by 3.1 y8 ~/ t2 o2 \* u5 ^$ O# ^
+ d4 ~' @4 A- ]8 @Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 35 _! B `7 j1 \* p
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 \3 W) Q+ S \Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
* T2 Y" x3 m+ `5 e7 ]( } = K^3 + 3K^2 + 2K( x3 h8 o1 u/ }! n; S. q
= ( K^3 – K) + ( 3K^2 + 3K)
1 o1 g" o0 \( Q0 c = ( K^3 – K) + 3 ( K^2 + K)7 }: K) g- H8 B n6 r, D
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( l8 T, M8 g$ J# O* }" f0 M
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
# c3 o6 P; y) p9 C& U$ A8 w+ F/ d' y = 3X + 3 ( K^2 + K)# x' W* V5 k) P# @3 n' l
= 3(X+ K^2 + K) which can be divided by 37 O6 a1 q+ r: M$ F1 o$ W
# r+ h, e$ q5 m$ XConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1./ m3 u h- B6 {5 s
" H; f6 [ K* O7 R( Q8 b+ F[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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