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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), i$ q) X1 x: n" F
e) \1 y! X6 R; zProof:
# b) g4 o; t- p- [4 `Let n >1 be an integer 8 Y" [6 @' g; p3 H% c8 O, I
Basis: (n=2)
4 W7 ^6 F7 e" O# Y* j0 _ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3# e; S/ E1 t r- P% Z% [: I( s
9 v; h& H7 U' q8 v4 ^" _/ W' }Induction Hypothesis: Let K >=2 be integers, support that* Y: f8 T& j# U9 r9 c
K^3 – K can by divided by 3.
# k- B& ^( l( Y0 R! p7 j8 y' @% M* ]8 [( ^* K& W2 F
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 33 f9 m, h7 f B/ W$ h8 |1 @" g
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem/ X; B/ x6 [& Z% ~; V- B
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1); k. a3 P' n5 k
= K^3 + 3K^2 + 2K
; a+ `% M# J1 j% |' ~, z; n2 g = ( K^3 – K) + ( 3K^2 + 3K)- f: x+ q( t4 e
= ( K^3 – K) + 3 ( K^2 + K)
+ c, ]9 x K p( d4 e2 Wby 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>06 o. C% ^4 p0 u( u& c4 p. M" x
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
% ^) X8 S4 @6 ~( e2 B/ N2 U( \1 h, S u = 3X + 3 ( K^2 + K)
% H' m2 J3 k5 \% p' g# u = 3(X+ K^2 + K) which can be divided by 3
4 }7 w8 X) m" |7 ^5 c: I/ V) T( d0 @" R9 y
) o+ i2 t. p: n; }% aConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
9 q) Y5 ^, h" m! L% z; g) C: h7 P, r7 n
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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