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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)% v8 b6 r! a6 ~
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Proof: " K4 }6 n. W8 d
Let n >1 be an integer 6 h0 e: p7 E: x
Basis: (n=2)
3 O# |! F! n6 `+ M% N' v 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
3 s4 S$ H% X# ]0 f+ _0 F
7 v* i, L2 i+ Y. S" Q: l2 n' ~. o7 O& TInduction Hypothesis: Let K >=2 be integers, support that
+ x; {( `7 w# u6 ^# G3 J# _5 G8 \ K^3 – K can by divided by 3.
+ ^/ H3 i! R% x4 e7 H0 f6 s$ p6 p% ?/ g; ~" |6 Z! D& X
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
6 l# c# M( T! G! X+ [since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
3 p$ J" j: ]% Z7 qThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1). v; d$ T/ C/ V- j7 l$ C: d2 N1 O0 g
= K^3 + 3K^2 + 2K
' t# q) H2 o a3 t = ( K^3 – K) + ( 3K^2 + 3K). P* U+ y5 V6 _; H. p9 ?. P
= ( K^3 – K) + 3 ( K^2 + K)& Q4 C* v7 H! D" r1 l; p/ ~
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>00 ^& p& ^) P: |; s; n
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)2 v* J9 e9 W. }: _+ S$ ~3 _
= 3X + 3 ( K^2 + K)
2 G. s8 A* ?& p' _% f = 3(X+ K^2 + K) which can be divided by 39 C8 W6 a* l) p- G/ u" [' s
5 U7 {. x4 t4 U5 d$ I; n- j
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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% K- u- S% q3 r# @ q4 Z[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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