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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 a/ N7 k6 E# i$ B, C; ^
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Proof: 7 h S8 U. J7 C& _
Let n >1 be an integer 1 o& t6 ^' K# B
Basis: (n=2)( ]( N/ V% C. H0 l8 G
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
9 |+ Y3 R: _) z9 m; A7 {8 r [! O/ C" G3 G
Induction Hypothesis: Let K >=2 be integers, support that o M: Z; ~1 R0 X
K^3 – K can by divided by 3.; T( ?1 f/ c! @( |% C: Q( ~: ^
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 32 S2 B7 \1 N) t0 g
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem: J, {3 t. V! `" ]
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1): i* G. {- j8 J( F) r$ c) [6 P
= K^3 + 3K^2 + 2K
2 u: o7 h" n- v. ]6 w+ ~ = ( K^3 – K) + ( 3K^2 + 3K)
# b% s9 o. O# x( t9 G = ( K^3 – K) + 3 ( K^2 + K)( H" }: i% I H/ D1 d
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/ R0 F U7 ]0 T6 y2 d1 {
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K). n d- I% b7 \6 @& y' z, q
= 3X + 3 ( K^2 + K)
. q- j/ p: y" O: N; @ p = 3(X+ K^2 + K) which can be divided by 3
4 j# o' I5 e% B; e+ s6 ~4 T( t# b
4 f j6 _4 m$ Z& b; Y8 IConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.0 L+ T$ I0 n0 ]
+ d% ?9 f5 \4 j, x2 a% Y[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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