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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)2 @& {; b/ A3 t. Q/ y
9 F0 \/ c7 m; l* n' M5 kProof: 1 y% D) f8 D4 P# I; b- x3 O) }: N
Let n >1 be an integer 8 o9 d: X" ~% j
Basis: (n=2)+ l$ F7 b8 @9 X8 y
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3 J9 F/ q* Q# S$ ]* S
3 D$ f' t; L, ]; D/ b7 CInduction Hypothesis: Let K >=2 be integers, support that
* V, j8 L# A8 x/ w! Y K^3 – K can by divided by 3.
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! n3 _9 f1 e. i# l; E5 E& R% k7 v9 |# RNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 U, {/ a) ~* i( Q- j- Dsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
. T: Y% R2 S! ]6 SThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, B' x- `4 f" M3 I# s' i0 Y = K^3 + 3K^2 + 2K
( d( [4 G$ ]5 v# q6 e. A = ( K^3 – K) + ( 3K^2 + 3K)
3 c9 e) T$ r8 \0 q* o = ( K^3 – K) + 3 ( K^2 + K)& R4 H& ?( @1 E- 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
d3 S* x& n! ?0 Y: r: E8 H. h/ L3 ~So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K) T& i6 z/ k3 I, J; a
= 3X + 3 ( K^2 + K)
+ x& O$ r: t7 G9 \& ^) L8 z = 3(X+ K^2 + K) which can be divided by 31 a0 w& M: F: ^. `' `; I
+ A! D2 x+ [' t2 [' `Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.6 ]1 H4 f) U7 X
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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