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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); [0 Z2 Y" _9 \: i3 ]
8 m* w9 c3 d2 ~' U3 ^$ kProof: 0 i( F' Q$ J! o
Let n >1 be an integer ) o. j: p$ \# n9 a: l
Basis: (n=2)
0 x5 ?/ R" O$ E# Z; g$ v 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that3 E4 `2 h, t- O, I( v! q5 e, o
K^3 – K can by divided by 3.
F9 K6 H: W! ]% {# b0 S
5 c4 b' j E" c$ P: KNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3# \3 J v+ O2 N0 r5 U
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem0 j/ q x4 ] @3 I
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)7 W; b$ X( G6 U" ^5 u
= K^3 + 3K^2 + 2K
4 Z: I/ \: ]) b' O% N) }1 `7 I% G = ( K^3 – K) + ( 3K^2 + 3K)0 Y6 U) P) C" U3 n4 }
= ( K^3 – K) + 3 ( K^2 + K)
: N/ |4 R! _. p7 p. ~! @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
7 k/ B) O% w; `So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 z# W9 U4 q8 d9 `+ W, K = 3X + 3 ( K^2 + K)
" o. ?; b4 A' p0 Q+ J' G = 3(X+ K^2 + K) which can be divided by 3
# I/ O+ M$ F' ^3 I% V+ K- _% w. Y1 U9 T
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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6 G) }8 j" }4 A" k: A[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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