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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)3 E) _' u" x3 H& T5 U( G4 q9 ]- q, E
+ O6 I* t, o3 K0 zProof: . d" T/ ]; k( u0 C# v
Let n >1 be an integer
Y& r& H$ `3 Z: X; C3 uBasis: (n=2)5 S! N) R1 l, e- G `' ]- A D
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 p& m- o) D _3 R4 s# ]/ P
* u4 Z2 I, a5 L: l9 I9 [
Induction Hypothesis: Let K >=2 be integers, support that
8 R) U& d c' i* h0 U K^3 – K can by divided by 3.
. O! Q0 A) U+ L+ t6 u' R
/ h" t2 D; M1 N6 Y4 p: z/ [Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 37 U% a, T+ s( V4 x2 |! \
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem8 |& j9 ^# _! N! q4 l4 c
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
' c2 R9 h( {) ]* g' m8 o = K^3 + 3K^2 + 2K3 A" @ F& ?+ Z# e% Y
= ( K^3 – K) + ( 3K^2 + 3K)
& P' c$ ~2 F: c& w9 U/ f = ( K^3 – K) + 3 ( K^2 + K)9 g- {! W0 l. @1 h
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; y f3 j" @, z: y
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K) O6 L- P- N8 M
= 3X + 3 ( K^2 + K)
9 Q8 n6 n( l- q' L; G% u4 G7 } = 3(X+ K^2 + K) which can be divided by 3
( o8 L- ]# Z6 e# t' c* `: b0 X9 b% X7 b; X8 R9 {- d
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1. j" q( H- }% Y! C% u9 p
; b1 r* J: u- F[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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