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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)+ [9 Z- D+ C2 u) \( X" }) P0 ^8 @
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Proof: 8 D7 D" g% V# D' D6 S
Let n >1 be an integer
/ r' F0 D. q- K* e. YBasis: (n=2). K' i& I$ M/ E" V/ C
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 q4 ~5 D% V3 S4 I E8 q
9 u9 l' v* E4 A, @Induction Hypothesis: Let K >=2 be integers, support that
5 h8 W. C' D7 ~+ \$ X K^3 – K can by divided by 3.- n! y/ P' B( y7 ?6 e" I# {4 U9 \5 f
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
# U6 H: B d% d0 _since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem0 J' W- C, `9 `& w5 _
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
3 b; R1 Y( H# n) n. ? = K^3 + 3K^2 + 2K- n2 L9 I6 A' U7 \ t1 i
= ( K^3 – K) + ( 3K^2 + 3K)
8 \$ ?/ t: x( w: q& w/ j = ( K^3 – K) + 3 ( K^2 + K)
. z6 E" c. f0 Jby 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
1 @! d4 S5 x+ \So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)$ c) I; d* b/ T' _1 T7 {# d
= 3X + 3 ( K^2 + K)/ y, Y$ M a# ~8 I7 C1 U% Y
= 3(X+ K^2 + K) which can be divided by 3
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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$ E& w5 Q: {# S" E! W[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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