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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) |# \' [+ o4 m: ^. o- E
0 s% {/ G8 n1 o+ ?0 qProof:
1 H- I; v5 E2 n E1 p; B6 d* LLet n >1 be an integer
7 i7 E- i! N& y2 ?# [! F+ kBasis: (n=2)
" H+ m9 ^- N9 G9 d" H$ B2 G 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3" ~- H6 u3 E T
3 g. W+ ]. U- `7 }- ?9 TInduction Hypothesis: Let K >=2 be integers, support that, {. E4 \) F2 x% e/ G4 a s
K^3 – K can by divided by 3.
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4 o- S, [' |" @& t. ]8 f, ANow, we need to show that ( K+1)^3 - ( K+1) can be divided by 35 {& _$ ]: t) x
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
- E9 G& D; c4 D* P! AThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
- v, o5 b0 q8 w ], o3 ^; l' m = K^3 + 3K^2 + 2K1 B( y! ?) Y7 J4 @4 E$ m# z! A
= ( K^3 – K) + ( 3K^2 + 3K)
7 q7 q$ s3 S4 F5 g3 i( |1 G = ( K^3 – K) + 3 ( K^2 + K)! J# c/ H. i2 C0 q& J. T
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>03 C" ^ T: f! r
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
3 a6 T2 f/ N( m6 | = 3X + 3 ( K^2 + K): }$ H5 j: k; D4 Q& j. a+ k7 c
= 3(X+ K^2 + K) which can be divided by 3& ~$ T6 g" b( M& m- @
+ J1 Q3 H4 l; B# CConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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