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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)1 G7 f1 {- [( k* X! r
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Proof: 3 n8 a% z+ y/ q2 E
Let n >1 be an integer " `6 D8 p) r0 k1 _4 n0 h5 @
Basis: (n=2)9 n' H2 B2 V1 N' m# ]1 L
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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2 ~2 K% V1 @) G# I$ `& @+ w, B5 a3 mInduction Hypothesis: Let K >=2 be integers, support that
) Y. [+ t4 f6 }; E# K/ a7 A6 ` K^3 – K can by divided by 3.1 U3 |" ]+ k' P9 h, t
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
/ W4 {5 g# w8 Y: b- f3 Nsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
" e! h* D( B. C7 l* Y% lThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)% @0 B- Y J2 t0 q+ ?/ Y
= K^3 + 3K^2 + 2K
# `. N- U* I: h- e. a% S7 P = ( K^3 – K) + ( 3K^2 + 3K)
" ]3 \7 X1 t4 P4 ] = ( K^3 – K) + 3 ( K^2 + K)8 N! H: Z* X4 E0 y( {/ K
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>05 h( U8 ?; [: o- |
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
) k2 s3 `! w/ z; @1 n# z, n = 3X + 3 ( K^2 + K)
# P5 W- j+ V& g7 k# m) i = 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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% h2 S0 o( e1 d3 y[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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