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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)+ v$ \. _+ L. ~7 W
$ \5 d: N- T8 w5 sProof:
( o9 v8 J1 N" i1 I! y. vLet n >1 be an integer 3 `6 s i) O7 f! }. @
Basis: (n=2)
" G6 P* y! Q: S: t6 B 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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0 z/ ]( Y L) {. J4 i% oInduction Hypothesis: Let K >=2 be integers, support that# B$ k1 E: \4 x7 a# N r$ ^! {/ I
K^3 – K can by divided by 3.8 d6 w6 j. h8 p1 ?% U
8 v. K$ B7 h- b+ e
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 32 ^1 y, G/ q- |2 j. ?
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem y7 F4 v7 v! }* N7 f
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, W) O% [7 Q, J9 x" [( { = K^3 + 3K^2 + 2K
; q( [' f4 u' A3 d = ( K^3 – K) + ( 3K^2 + 3K)
U1 a1 G! A- b, U = ( K^3 – K) + 3 ( K^2 + K)+ T U! Y- B' }$ 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>09 O( d S$ I2 s, o4 f
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
: k) n3 j( v. p y = 3X + 3 ( K^2 + K)
3 e: c# L5 ?9 k3 }+ N9 F! V4 M = 3(X+ K^2 + K) which can be divided by 3
8 B8 J5 W7 u$ D3 f
+ p8 H6 p" J$ o/ i& YConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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; G) `: l* E4 F* _$ M4 V1 y[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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