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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: ! e1 C9 C/ N0 L2 \; N; }+ y; Y
Let n >1 be an integer % M7 w2 a2 y5 j. u- I+ Y
Basis: (n=2)
3 ?. u, @0 }" F0 _ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3! i' d3 \1 b9 C7 d. z O
2 g- N/ P( Z/ rInduction Hypothesis: Let K >=2 be integers, support that& A( W0 E. ?: u1 ?" s! @( ~
K^3 – K can by divided by 3.
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5 O5 Y1 G! x5 f N/ WNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 35 K# z5 r' c U% Y4 {
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
: K+ Q7 ?3 Q$ C; LThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' o+ f* {8 ~ D4 v9 ]4 D2 Y0 S6 A
= K^3 + 3K^2 + 2K
3 c7 \, O8 Z" Q6 M8 j = ( K^3 – K) + ( 3K^2 + 3K)
; ]* Y8 i- B/ i; s: l( L; [ = ( K^3 – K) + 3 ( K^2 + K)# P# h; P5 L8 Z3 |% l
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
* ]/ X9 x+ Q' o1 ~" DSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ m# h% a/ T+ a- E# {5 a7 g = 3X + 3 ( K^2 + K)8 N8 m# d6 l5 x: f: w
= 3(X+ K^2 + K) which can be divided by 3
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( n; Y* a2 P4 A9 LConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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3 n$ f" k N1 u7 U# H0 o' q, X[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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