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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:
; F- ?. f& N$ V6 c8 E; GLet n >1 be an integer
0 ? g% f! I; s7 DBasis: (n=2) m3 i3 Y% z3 d0 D, t- S
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 34 C& n8 q% ^4 ]# l$ T2 t) ~& O
2 }# ]) D$ ?# W3 d4 I, B0 F- B2 o0 uInduction Hypothesis: Let K >=2 be integers, support that3 h9 q+ T5 A& g- F9 a
K^3 – K can by divided by 3.5 z8 V" q2 \/ B( @. c+ S
4 c6 b( E: S" q h' rNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3& V1 h! \3 k8 ]1 ]+ [
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem+ f" p6 k0 ]5 x5 f* X5 _
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
4 z* `9 x( v' F7 | p = K^3 + 3K^2 + 2K
4 v2 x+ l# i' ]/ I! M4 m = ( K^3 – K) + ( 3K^2 + 3K)
7 d0 r6 E, j R2 x = ( K^3 – K) + 3 ( K^2 + K)
, B& B# }. L( r( m$ G7 aby 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& V$ x- ?; U. _8 R5 W
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K); X; x" \7 [/ {, ~! M
= 3X + 3 ( K^2 + K)
1 F! C% a' V0 G; u4 p = 3(X+ K^2 + K) which can be divided by 3
1 |1 a8 k+ u9 K, |. `& g8 j2 Y- Z3 c2 H4 T: y
Conclusion: 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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