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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: ( t* V( @/ \; p! E* [( L& @# J
Let n >1 be an integer # T! T. n3 p! b: g
Basis: (n=2)
! n3 A* O2 P2 ?8 }+ { 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 e( m+ z% y P6 o1 d) s
! K9 h0 _ c" j' LInduction Hypothesis: Let K >=2 be integers, support that o8 V5 \& N9 q+ P7 C5 c
K^3 – K can by divided by 3.
T. t6 S0 v9 v, C1 G Z- N) L0 J2 B$ R# e) L1 ]; @$ o6 N
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
! s) }6 W2 x. m+ f: A: }! J7 I: K( xsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, v7 \. A2 P! v gThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
/ [ q. _0 r7 @7 Q/ { = K^3 + 3K^2 + 2K
# s. `0 f, Y- X: i% w2 M = ( K^3 – K) + ( 3K^2 + 3K)
- ]- X; t$ L3 _1 c = ( K^3 – K) + 3 ( K^2 + K)
, E2 q$ R7 } m7 d9 ?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- W, W$ T4 B' E- `; }" x
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 y, F* Y6 U3 \4 ] = 3X + 3 ( K^2 + K)
: A" X' \6 E" d4 ~" T = 3(X+ K^2 + K) which can be divided by 3" k* v/ W# T' o* A" }5 _5 {) A8 L
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.8 G/ N* H% @8 R6 Y" _
% y0 z( h' g( q4 \8 L[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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