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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)6 o2 z2 |) \5 a
; w: f9 ^2 u' F% |Proof: & C1 g* @( a; H( x) j
Let n >1 be an integer
: c( t' V: s0 l& F$ oBasis: (n=2)
2 ]. V, [1 A% p: o2 m/ [ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3) C# r& w/ T% p* u$ U& s5 I; [
) X( W) v& W' O- B, Q2 K; ~Induction Hypothesis: Let K >=2 be integers, support that+ O2 k9 U/ f% y6 L H% B
K^3 – K can by divided by 3.2 M0 M6 Z% j; q; A' u; _2 v9 x
6 F+ L7 |. {) L. |# N$ r) rNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
% U$ J- O2 W5 B1 u7 V' B1 fsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
8 F4 d: ?3 A/ R d4 |Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, Q7 P1 ?: k7 _7 c = K^3 + 3K^2 + 2K
! E9 E8 A: E( J; W; f4 z. g = ( K^3 – K) + ( 3K^2 + 3K)
/ {9 C+ P# d5 j* M = ( K^3 – K) + 3 ( K^2 + K)" h1 C; l, [7 [8 [) m) K2 N: v3 c
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>02 r* L+ A6 s b% E: H
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 {6 p% k. z+ n G! [$ j0 p = 3X + 3 ( K^2 + K)
7 W/ k% {( s$ ^ = 3(X+ K^2 + K) which can be divided by 39 s3 {, l7 R4 `: ^
- M2 i% y7 z% c; _* J$ AConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
5 N1 @, q4 F; Z3 e' Z1 c- C3 r2 n$ Q. t' j% q/ E# g
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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