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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 a% {* u2 a: [! G& e' S
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Proof:
) I( \- X/ a M+ p( B4 i1 j BLet n >1 be an integer
/ D- f/ ~1 [) v- j( M8 c" a# QBasis: (n=2)
1 o2 C" W) g/ ^4 P% i 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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! }7 Z6 E3 h# ?# ]4 t: PInduction Hypothesis: Let K >=2 be integers, support that6 Q" ?) x/ h* `, x* v
K^3 – K can by divided by 3.
8 d( K) r4 M8 Y9 s2 B$ I6 _" Z a/ p3 [5 V; ~2 }! H; {
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
+ ^- J& F, O+ Psince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem+ ~8 n% v" ]$ ~
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)$ P E( \* R5 M9 m5 ]! V
= K^3 + 3K^2 + 2K
B* r/ N1 j3 |& p4 ^ = ( K^3 – K) + ( 3K^2 + 3K)5 M1 \$ S! I5 H( e" @4 ]2 ]
= ( K^3 – K) + 3 ( K^2 + K)$ d$ u; D7 w$ g6 u) m
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
# z* ^1 X7 F$ O4 xSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
5 u' \+ s" n. v9 V" N! ]* } = 3X + 3 ( K^2 + K)
! v/ M" B! P& b = 3(X+ K^2 + K) which can be divided by 3- A7 t- B' e! w1 s0 G3 Y
4 \$ f) D% E$ P) k! P0 e# E( iConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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