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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), }+ T C2 i+ D0 l: x
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Proof:
4 j/ L8 w4 T5 m6 ]1 ~Let n >1 be an integer
! {# o8 ?( y6 i2 V: y2 u3 W% }Basis: (n=2)
$ q% u* l- w% U- ]' {+ X, \ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 34 P* b$ n" O/ q- w. {7 X
/ v! M* B' z+ ~3 _, a" ~- nInduction Hypothesis: Let K >=2 be integers, support that! s0 M3 ~9 l7 O! B
K^3 – K can by divided by 3./ e: n z+ w9 N9 P9 U
/ Y u5 k: z5 vNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
: K) B: m& s9 n7 Z/ F" G ^since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
1 `! x" ]# W2 zThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
0 A$ z, v" l2 m ^+ n = K^3 + 3K^2 + 2K
& q* X% a3 w( ~ = ( K^3 – K) + ( 3K^2 + 3K) a! N- J7 D1 ^ {& v; s" w
= ( K^3 – K) + 3 ( K^2 + K)
. o2 C) v; ~- e/ `, U/ F3 S$ Kby 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) |7 x% `6 Z; T, B8 ^
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
; O% V1 A: X9 g% [ = 3X + 3 ( K^2 + K)$ [8 h8 T" X* b* l/ O2 n
= 3(X+ K^2 + K) which can be divided by 38 {' }0 z2 D1 o; J
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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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+ b1 U Q+ k+ U" |[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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