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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)9 ~! d% N5 T0 X' `$ M" W
7 h& @5 I4 v r. [* n/ x/ b" R& ?Proof: 7 h. m; C: O# o
Let n >1 be an integer ( q* F2 \3 I3 P) ^
Basis: (n=2)$ Q. k( s0 g3 F7 g- X! r6 X4 d' s% N
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 that, H9 m, C8 i" J5 \! ~
K^3 – K can by divided by 3.
4 F% p/ D" r$ ^9 [. a8 f; z: o8 U6 g! v% }$ J) X, P( W7 L6 [& h! v+ X
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
' K; u; ~# x) Wsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
" n$ Z. l( A) u0 t _- t. e; K7 nThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)7 w( P: L3 x# u; }2 S& o
= K^3 + 3K^2 + 2K! ~8 |- P- i9 u/ N
= ( K^3 – K) + ( 3K^2 + 3K)( I; s- f) s# @- P
= ( K^3 – K) + 3 ( K^2 + K); b' m! o% R0 M/ i$ A7 _ p( m6 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
' w: k. y$ k2 O1 h0 G7 y* k( A P, ESo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
9 Z1 y/ a2 j4 b) \ = 3X + 3 ( K^2 + K)4 R; H O: G: \0 }2 G, V& o7 a
= 3(X+ K^2 + K) which can be divided by 3, Y$ M0 T& v! g3 R- A/ B1 t
' y+ Y7 C- H6 e- r( K/ ~% }
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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4 X; X- O6 s! @# E2 K( v# H[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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