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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)8 s/ y( V) [* U; U+ B! p: H0 U, ?+ h
3 e- n1 B2 s0 t/ HProof:
: f! F1 F5 P5 A+ A) C% w' DLet n >1 be an integer
; S* E- |0 p+ {7 O j8 m* p* MBasis: (n=2)
: _+ T; x! N7 b! ~/ l 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
; b+ W) W+ z" Y6 p2 W% D5 g7 `2 a9 @- }! _
Induction Hypothesis: Let K >=2 be integers, support that
. t4 B! ?4 \) K i0 ?( a/ k K^3 – K can by divided by 3.
) k4 \7 {/ X$ q% V( h3 B
5 E. r8 J% ~0 t7 N; S4 t T- U, ^Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
6 R. A- t. P8 m, M U% Xsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
4 V. N8 `# N) L8 l; E _Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. r6 D8 n/ e8 f( h; f7 a! c = K^3 + 3K^2 + 2K
8 X& t* C, o; l; X& X+ I = ( K^3 – K) + ( 3K^2 + 3K)( S; Q/ K: R$ y1 h1 [
= ( K^3 – K) + 3 ( K^2 + K)
5 R) }+ A# c) }. A1 L4 Nby 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
& I Z" Z y7 z" o2 H: B) H2 _% \' m& ySo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)9 P" [" g# G# N
= 3X + 3 ( K^2 + K)
" A/ F& \0 [8 N2 l, a = 3(X+ K^2 + K) which can be divided by 30 C0 Z; k+ x- \
6 e" k4 l( v! x0 ]- p* Q
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.) ~5 L8 ]* Q9 s, e _0 Q- l
+ z: G/ r+ P1 p4 w. O[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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