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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)9 B0 Y9 L- Z5 l
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Proof: 4 P, v% c* \8 A1 K
Let n >1 be an integer
9 g# \% |& Z! f; n) b) JBasis: (n=2)
1 t2 I$ \& r1 Z6 O3 o 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3( }: k" G: {! H3 `
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Induction Hypothesis: Let K >=2 be integers, support that) j( L0 o. y/ A. o; T
K^3 – K can by divided by 3.2 v- S' U5 Z$ z: |3 w; Y: r: R- P
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3+ o) _) y0 f/ M9 P u5 D
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem1 h' [' { a0 c
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
: A% R2 G0 z! ^; m) y = K^3 + 3K^2 + 2K; `% X# b# F) N# w9 c$ |
= ( K^3 – K) + ( 3K^2 + 3K)
1 i4 w* o( ^! o+ e5 P5 D = ( K^3 – K) + 3 ( K^2 + K)8 X* c- l8 T& G
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>0
0 ^8 n/ {/ X; S+ xSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ o& B7 S k8 m7 y$ o
= 3X + 3 ( K^2 + K)0 o. Q% p7 ]. g8 C9 v) R6 j
= 3(X+ K^2 + K) which can be divided by 3/ {0 Q! r, c, }3 `5 w) k# k
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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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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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