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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: 4 Q/ a3 S7 x) M; c A* ^
Let n >1 be an integer
\$ V) a' B& B3 f# t" n" yBasis: (n=2)5 j' K4 ]3 y4 L \) x
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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' I$ }5 e' J. T" }" T5 @Induction Hypothesis: Let K >=2 be integers, support that7 N! \/ s0 Z8 P, b) `8 q; x
K^3 – K can by divided by 3.
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4 p* Q) z% u% X& l8 {! fNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
/ c2 r' k0 h% ^( W7 e, e/ ksince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
# G+ P0 i3 ?$ G' {5 q2 o/ tThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
; ?$ E" a g( A = K^3 + 3K^2 + 2K
6 Q6 c0 `. m7 O: { t* U) R( p' w+ e = ( K^3 – K) + ( 3K^2 + 3K)1 D9 y" T3 w: x& @8 x
= ( K^3 – K) + 3 ( K^2 + K)
2 q4 n2 j) F. `0 x2 `# \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>06 D: D) ?' s/ c5 H" k
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)# @3 I7 t! K) B4 A. Q* F
= 3X + 3 ( K^2 + K) I% b z6 y8 b( K; J# O* M* ?- \
= 3(X+ K^2 + K) which can be divided by 30 m# [: I$ \1 |: |# p
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.! X3 x# O2 Q* q" z; T
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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