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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); n, R& W3 W: Y% X
# t7 c. |2 e: g0 \% {# d$ TProof:
4 C3 U8 k8 E% f( YLet n >1 be an integer # r0 u+ g; l. o6 \
Basis: (n=2)
# y/ | Z' V! R1 E' N1 F 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
/ _+ T, k% g& u k3 u, t
$ R" ~% X" h% i( A fInduction Hypothesis: Let K >=2 be integers, support that
% R* D" Z! p! ]# v, z5 w8 w K^3 – K can by divided by 3.& z: f- b- k; i% h7 U8 g3 }
& f) N& h* w% z; ?9 tNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 39 b' Y9 f" B5 I; s3 W; h
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, m# Q& F f7 B' }! z, aThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)( x1 n q# ~: V& @3 ]; l( t! |; w, z
= K^3 + 3K^2 + 2K
" h7 A0 h' Y: p% X = ( K^3 – K) + ( 3K^2 + 3K)
3 I; w6 h' Y: L) K5 Z4 p$ L = ( K^3 – K) + 3 ( K^2 + K)# `+ A; q$ ]" c" K$ q
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
6 z. f/ C. i8 t+ ESo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)% L% V# a- r0 D9 c
= 3X + 3 ( K^2 + K)8 d A5 I, g( m5 {+ A* Q* ]
= 3(X+ K^2 + K) which can be divided by 3) q7 J$ D* m3 ^* L
0 T. J* Q: s3 J( {4 MConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.5 n% q* c# ^) P
O" R3 R: ]' N. B
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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