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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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Proof: - e' V R9 [: u2 I1 c
Let n >1 be an integer ' n2 Y0 e3 E0 s" y
Basis: (n=2)
7 i: t7 S, \8 `: T* K! G; d 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
- j0 E- \1 n- n$ \) C+ ^9 K" j: s- P$ o" b1 r7 J
Induction Hypothesis: Let K >=2 be integers, support that
, |& A6 v$ x, A/ a K^3 – K can by divided by 3.
/ n! a. k, _2 i, r2 {: b, k+ z4 y5 [( g# q! U6 S' r
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
7 I% [# N& v% c8 S, ~' Dsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, f$ a2 j1 N: k) r" @6 i1 p" ?5 U, w
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)7 A# O" {3 J" L: B4 ~; g
= K^3 + 3K^2 + 2K/ ]' a1 U8 `! g- E h2 Z
= ( K^3 – K) + ( 3K^2 + 3K)
6 q' X& r! s: k6 ~. i2 r; k = ( K^3 – K) + 3 ( K^2 + K)4 P0 X" a) M( J' {0 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. k1 H1 ~- o! J& N8 ^* B
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
: `+ `: n- M6 G! |8 x4 e = 3X + 3 ( K^2 + K)
2 c6 r6 }) y8 |+ s R = 3(X+ K^2 + K) which can be divided by 3$ r' Y2 B& o; a: q: \ o
9 {% O, p. z+ K2 |* |2 Q3 m; hConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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3 U) M9 p% k2 ?( e# I. r) W[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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