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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:
5 x) ^' m# X9 o+ ^' F6 w/ PLet n >1 be an integer
* Z/ t6 X$ Y4 b( [$ v/ vBasis: (n=2) [) u* L) S7 C' ^
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3$ q4 `: n* e. C
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Induction Hypothesis: Let K >=2 be integers, support that
0 t# n- n7 j- ]( \ 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 32 N! U, d; q( h
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
; q" D& K) n, e3 _0 V# hThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
) @' I" J. q2 ^$ H = K^3 + 3K^2 + 2K- h5 i* E' L+ S/ f
= ( K^3 – K) + ( 3K^2 + 3K)) k3 M7 [ g% `5 B2 {. [" w
= ( K^3 – K) + 3 ( K^2 + K)
3 M% B* l5 A2 J$ X3 nby 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
' o- ?9 p5 F) c' _: w9 ESo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)9 K" l3 A' l4 z- Y- H2 V$ s# b" T
= 3X + 3 ( K^2 + K)4 R0 ~" ]. F- K6 O# N4 G I. m
= 3(X+ K^2 + K) which can be divided by 3
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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6 Y, I6 U" T: L4 }5 Q% H+ D[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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