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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)
h- L" c3 o, ?9 J$ O, i- X, h3 {9 ]4 I/ w6 ?
Proof:
. ^$ d4 E6 H7 ^8 f) w0 iLet n >1 be an integer 0 f% T3 T5 V. X: o3 V
Basis: (n=2)
" }* W2 L; Z4 V( a: w! D 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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; A/ e$ |5 }% q7 j; L6 oInduction Hypothesis: Let K >=2 be integers, support that9 x+ d b: X; q H4 u
K^3 – K can by divided by 3.
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3 E6 [$ A) _9 F& ^Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
8 @! z* R r p: w% {since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, ?9 a; `0 ~$ P
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
* ^: T& c: `" W0 M = K^3 + 3K^2 + 2K
; C. U" ~7 [" i# B1 \8 E = ( K^3 – K) + ( 3K^2 + 3K)
, x6 r2 j" P0 m' g. i0 m = ( K^3 – K) + 3 ( K^2 + K)3 L8 p ]/ }) A1 k/ i2 W& a/ B
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% q& R2 _5 q! g
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)5 e. [ j3 x* M) t" }& U0 O
= 3X + 3 ( K^2 + K): W4 Q) h) Q4 A/ j( f' r! D, W
= 3(X+ K^2 + K) which can be divided by 3
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3 ~: w# }! H: t* A$ A% B4 zConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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8 N" ~1 W7 c+ T/ A3 {0 V[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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