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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)+ t; b" F( f, I
9 J" ~, j3 ^5 g5 U5 M
Proof: * N/ _% E3 F" G$ G) `1 ?- f
Let n >1 be an integer 1 S% \. T( [7 I' O8 ?) o3 `
Basis: (n=2)
' k+ `0 K5 d* u: L2 p 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 35 ^8 c$ }( ^& k" g& U
1 N/ t1 I$ a. x% x7 D% S0 ^8 A: T
Induction Hypothesis: Let K >=2 be integers, support that
; H5 |' ~* w, G+ s% p$ Q' j: F K^3 – K can by divided by 3.0 {4 \; b! v5 ^+ c9 a
, \. {0 ]* n: V) T% KNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 35 s; a$ x7 R7 D2 J/ H' p
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
& z1 Z" A4 R) p6 UThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
2 K9 Q7 n# e8 T1 k+ ^; C = K^3 + 3K^2 + 2K
8 ]0 l1 k. T& j& _ = ( K^3 – K) + ( 3K^2 + 3K)
+ A7 G* f; w' R* S4 Y* ^: J7 r = ( K^3 – K) + 3 ( K^2 + K)4 ~ P! L- \* X/ |$ i& _ x! w4 \# S g
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% ^- s3 R5 t3 }! C/ M6 p7 B
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
. S3 h2 }- w0 `: T U6 N9 z = 3X + 3 ( K^2 + K)
7 |5 x7 t% L- ?. L6 r" u* L7 L% \$ I = 3(X+ K^2 + K) which can be divided by 3
) c8 [6 v1 m8 c1 Y) q/ d$ Z
% z+ X6 t3 a$ }0 D; k9 m; m5 rConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.1 D: w8 I- g9 |: `
1 A* L) ~ V j
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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