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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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# v$ k3 a+ f# C* P* dProof: 2 \+ X4 B2 y( h& y' {" o8 z
Let n >1 be an integer " y- E: W' [: d* @4 ^! O" k$ _
Basis: (n=2)
2 u6 H; S+ w# r W3 u4 T 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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- B- Q( I+ X9 n/ U) r* w7 AInduction Hypothesis: Let K >=2 be integers, support that
. o7 x6 `+ s p/ B& ~8 u. K K^3 – K can by divided by 3.: O, j# l9 y2 k* z+ h
0 I/ j! C w; d4 e2 d" L# QNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 v( Y3 Z& }# B- |7 s& W# p; `since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem' R+ _/ l; v4 X' \0 M
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
- `8 l: {6 G1 J1 ]( c3 I" { = K^3 + 3K^2 + 2K
, a% G9 p( z+ | s, F( h; t = ( K^3 – K) + ( 3K^2 + 3K)# }8 V+ A6 l( Y7 @
= ( K^3 – K) + 3 ( K^2 + K)# M7 M3 I1 \ H% f0 ~6 {' a9 J
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! A- @* d+ y8 I# U0 b
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
: N3 k" Q4 b! x9 B5 o2 x& A1 s = 3X + 3 ( K^2 + K)
, a- R5 Y1 n/ e# c = 3(X+ K^2 + K) which can be divided by 3
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. x( V8 ]: ?8 h. w5 h% {Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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9 ]; `8 ~% }" Q; O: B[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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