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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)7 ?; K4 ^: a2 C0 g7 a$ \5 B9 }0 P
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Proof: 6 h- j; u# G) n$ \* y/ V
Let n >1 be an integer & N: R& J) A) J! G% P
Basis: (n=2) Y$ @ `! P, C* r" \. _
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
2 w% S5 O/ y; x b0 N, ^- ^. R) a: x' x6 G, v0 N/ |6 q: A5 Y. E0 q
Induction Hypothesis: Let K >=2 be integers, support that
7 G2 p2 _/ V1 @ K^3 – K can by divided by 3.' g9 O% @3 i6 H& H+ R" B
( D( e- l- A: x d" u" M; JNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 37 Q' U# y+ x5 [* N" \
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
( b3 i0 b9 e* r+ [/ LThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)" G6 O I! D+ Y1 P
= K^3 + 3K^2 + 2K
7 _' y) {6 m2 {8 [! H2 k = ( K^3 – K) + ( 3K^2 + 3K)
! p# o6 t+ Z/ }. T& s; {2 S = ( K^3 – K) + 3 ( K^2 + K)8 f0 D( _4 m- e# {1 E" n% O
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
3 l; S9 l. }' i S) V7 mSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
- p3 j: h* y, D$ j = 3X + 3 ( K^2 + K)) x- O3 A0 e# y: t
= 3(X+ K^2 + K) which can be divided by 37 R* J' v$ }# a! K; ?
8 B3 b: b6 n" b |# ~Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1., b# C7 b1 o& J) U/ q6 h
2 ?. k+ ~ X# V$ _. ~+ w( s[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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