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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)& t: S( M+ {$ |3 T- }" T/ p
! l+ Y% S2 G; h% _Proof: * }3 P! R1 Y' o) `" J* c# L5 B
Let n >1 be an integer ( y+ S/ D( l8 _) t5 L, U& i
Basis: (n=2)
7 k* ?4 S! t+ |! x3 ^" v 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 p) \! \; l4 I6 [2 W# L+ ^
! | ]% ]% M' U: }5 X* EInduction Hypothesis: Let K >=2 be integers, support that4 i c5 e+ \! B/ W# y
K^3 – K can by divided by 3.: R, o& V# l' G3 R6 }! K
% _- i0 u& n2 _' Y2 ` d& N
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 34 p- L! o% G. m% h Q( j
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
& P. i1 `( i' HThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' n- S; _8 n# }" f7 Z( W* z2 ]; v( G* x
= K^3 + 3K^2 + 2K: X1 d! m; V# I. u6 N% s2 c
= ( K^3 – K) + ( 3K^2 + 3K)
7 p+ Q; M' q) b- E# { V8 S = ( K^3 – K) + 3 ( K^2 + K)6 w7 X: D4 d( q/ \+ k6 y: ]( f4 d
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: D7 o8 u' c1 n
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
: U( o$ A! ?6 H: \" p1 B = 3X + 3 ( K^2 + K)( l: C% F0 D8 q, U
= 3(X+ K^2 + K) which can be divided by 3- h! g) w% X/ y$ M8 x F0 @
9 t1 Y9 J% P. b, D
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.8 ^2 w) K( o+ I. a
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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