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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)4 p3 A0 B. R9 g
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Proof: b) z0 R) C' C9 u9 W! S* A. o
Let n >1 be an integer 0 i% {7 X$ ?! O) G$ \$ Y' s
Basis: (n=2)
/ p+ T. I" ]7 I2 q5 a+ g( | 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 30 @) g' g6 a. @. g9 _8 b; N" K1 H8 f
9 a2 A5 ~8 j: ~, g1 Q% [/ tInduction Hypothesis: Let K >=2 be integers, support that
G) P1 P& b! W, r K^3 – K can by divided by 3.( M% ?, m. L- G) i
" X! Y$ E' b! d1 lNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3. q0 y% U: Y4 M; y6 p
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
. `! d$ E: P: ]7 h, T, k# ~7 a4 w; D; GThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)) M; n$ _" D" m" B3 f4 @
= K^3 + 3K^2 + 2K8 L6 ]& U& n) b K0 u
= ( K^3 – K) + ( 3K^2 + 3K)8 ?( V3 Q* p* K! Q* C: \+ ^8 r& G
= ( K^3 – K) + 3 ( K^2 + K)6 f; n6 A" [' X! V0 _8 V$ K+ C# A: i7 {
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
; W: T, S: L7 L/ R. n ^2 PSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)4 s: k& J$ M9 W! i" E6 E" i
= 3X + 3 ( K^2 + K)$ Z7 B% }; j" l1 k2 w
= 3(X+ K^2 + K) which can be divided by 34 }" T2 D7 j1 S/ B5 O. `/ J
5 T$ X. R7 U/ S0 E$ U( Z/ kConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.% E% A; r& A O, b" w
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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