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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)4 J H5 P/ W& R9 V% |: D; w' h$ i
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Proof: ' ~) b. L, R* P! ~- r9 k6 ~/ A
Let n >1 be an integer ! a) p( C; a+ \
Basis: (n=2)$ J# F5 _9 `8 e& E. }( [
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 38 E$ C0 O7 }& d: H
( V" v9 o% P6 o& `7 G W/ jInduction Hypothesis: Let K >=2 be integers, support that
9 \# r( d. H/ e2 k K^3 – K can by divided by 3.
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7 z$ H: I( [6 F) v | QNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
2 a: e( C) W1 `1 H* D9 [5 ^3 |: Lsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem' L+ k! K+ H* H
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1); G7 k5 s& `! P2 F
= K^3 + 3K^2 + 2K+ x3 Q" u \$ `5 w" x
= ( K^3 – K) + ( 3K^2 + 3K)+ U" c% I6 p5 M4 r0 x( ]
= ( K^3 – K) + 3 ( K^2 + K)
- s/ S% E' N$ z% t+ yby 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$ j# h0 O3 k* M/ X y
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" Q# k. G: b j, J8 s# q; h! P = 3X + 3 ( K^2 + K)
3 O# Y* j6 S. P = 3(X+ K^2 + K) which can be divided by 30 ^3 ^( t" H+ i* K& S% E4 Y
4 V n- o8 j+ P! ]( d( y8 R: BConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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, E5 K5 M T7 F" \1 U[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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