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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)$ K9 e( r B3 ]3 T) x- ?' |, N
3 X8 `& B2 K" _9 }- j& L2 ^Proof:
5 N! P5 V5 ^6 QLet n >1 be an integer
# E5 ]) t/ G; P+ tBasis: (n=2)
2 O3 j8 o8 K! i 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
( ?4 ^3 g- I3 L% F4 x& n* ~) ~1 P K^3 – K can by divided by 3.0 b3 t) H% [% f- ~! E
% b1 }# d6 u# n" n# S4 mNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3- w v: F- n2 G5 f& S- L. m( y
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
8 B6 H1 w$ m* y5 HThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1), v5 I4 ` E9 c$ ?: E! W
= K^3 + 3K^2 + 2K
5 |! v" G( q7 s- w; U S' Z = ( K^3 – K) + ( 3K^2 + 3K)
' Q5 }' T- \6 { = ( K^3 – K) + 3 ( K^2 + K)
% k7 D" T0 k' `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
7 h0 V! r/ n1 G) v A; rSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
' P: ?7 v7 W/ N; l# ]; w6 U- p% t = 3X + 3 ( K^2 + K)
8 D2 b( o1 O+ Z8 U = 3(X+ K^2 + K) which can be divided by 3# ~6 S# l* a3 `) O7 A- `6 Q
6 p! C1 p; d" p2 \- i) L5 NConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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. O5 ~$ k, U6 v9 s6 X[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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