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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: 0 d% k" i9 P. A+ c3 a
Let n >1 be an integer % a8 ?& n2 O8 `( K' [- k
Basis: (n=2); o7 X0 F! k) U- ^. B j+ C
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 G+ k8 l/ i& g- L: J$ @5 r2 a, y
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Induction Hypothesis: Let K >=2 be integers, support that4 y: y" a# ~6 h b$ |5 N) m U
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 34 W: G1 q. }4 G8 e u
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, q$ a3 x" b) G
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)/ T5 O( F& F' _% E3 H! E2 }
= K^3 + 3K^2 + 2K7 q( i+ ~) {" g4 K7 Y5 ?
= ( K^3 – K) + ( 3K^2 + 3K); `5 C, `4 F+ z, J7 r
= ( K^3 – K) + 3 ( K^2 + K)
$ A! e4 T0 P: ]/ Sby 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
9 d( [* o. S; b. g' s( Q. |& RSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)+ h& y) U' p! {1 w0 \
= 3X + 3 ( K^2 + K)2 J0 N- R3 u( Q$ z0 S8 H
= 3(X+ K^2 + K) which can be divided by 3* t( Z' U0 W" S7 D9 f( {
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.; u3 R1 F5 d" o8 J: f
; s; W/ L! T+ y' B5 g% V[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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