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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)6 v3 z7 R3 @4 y( L$ k& O5 G
/ `7 A1 H, ~8 x/ ?. x- t# H. JProof: 1 A, e1 g/ e' I3 R" c/ \ {
Let n >1 be an integer 8 l0 a- L/ Y0 t. o9 S
Basis: (n=2)5 n( K c6 O% f" R( Q% K+ K; G& M
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
6 {" C! t! Z0 r4 {1 g6 i0 u7 w3 E+ m X9 h+ y% f
Induction Hypothesis: Let K >=2 be integers, support that
% G7 y( P- X# C5 t& ~ K^3 – K can by divided by 3.8 y8 F T% n. [* T' ^; M
* E( ]2 o( K8 X, v" e0 s/ G
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
( d4 b5 G4 C! Xsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem1 Y7 a G; ]% C/ I& c$ B
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)# o: u+ T2 A( n, R' d* w8 a
= K^3 + 3K^2 + 2K/ q {% C' E# [1 x
= ( K^3 – K) + ( 3K^2 + 3K)- ]! d7 h7 b- a5 b% g
= ( K^3 – K) + 3 ( K^2 + K)
. C. g; H: i# B8 l \/ q) w, ~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 `2 r2 D$ g: QSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ Q$ D( M# l9 Y& ^ = 3X + 3 ( K^2 + K)
x# L% K2 ?. _* A F! ` = 3(X+ K^2 + K) which can be divided by 3) i* I; O1 c* f
S% ?' F- w( c7 Z9 U, {- p
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1. L7 b1 o& G: p9 {
' |5 h# J$ J0 z[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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