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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)
* G' O! u2 t( y8 q4 G' _, o1 V; D: C& K* H! a# z+ {
Proof: 9 R; _$ r, g. `
Let n >1 be an integer 1 T: H u4 \2 w7 V' a/ w/ b
Basis: (n=2). R# g2 _3 q) I s4 |5 C# f
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3$ l3 |2 k6 k {; U
. H; M; J# I+ C% {6 U' uInduction Hypothesis: Let K >=2 be integers, support that% J% ~# c1 {, V: `* h- Z* w6 K
K^3 – K can by divided by 3.7 N. J% e- h& e4 W% o' b4 e W: X
, z0 w6 ?/ ~* m5 S! \Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3' t6 B3 b$ @; g( F6 Q& |
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem* H4 o2 S, K, ~7 r6 U1 ^) a
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)- a1 }$ D, ]" Z" F. z! _6 \
= K^3 + 3K^2 + 2K+ b* x& p" O# W+ |
= ( K^3 – K) + ( 3K^2 + 3K)
, ]/ \8 N" a, Q. x- E% O4 g, z8 | = ( K^3 – K) + 3 ( K^2 + K)+ B3 C0 m' \- k$ d# a
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- L4 G$ O! U% ^: O- x
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
' C' e2 l/ f* `; H& d = 3X + 3 ( K^2 + K)
4 D) t) e R* r3 q% P4 @ = 3(X+ K^2 + K) which can be divided by 3
& W9 [. U2 [8 g! ]3 a: c
+ U& t3 A+ k$ `; H$ jConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.9 ?7 \1 p9 L6 F4 P0 l7 ~
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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