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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)" D7 K' y& j* e% ?1 c
7 t- E, Q; ?0 f! ~5 KProof: 7 C: ^: \' @$ m
Let n >1 be an integer
( b' S+ a7 K' BBasis: (n=2)
* I$ [% P% I a6 r+ o2 H3 q2 J 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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0 J- o) q& K @0 f5 EInduction Hypothesis: Let K >=2 be integers, support that9 g; J* H: e, e/ b8 q2 l
K^3 – K can by divided by 3.: N/ j J5 f! d5 \
9 M$ U7 |8 c/ w. q: E2 w$ y
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
5 q+ A! Y5 e L6 msince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
1 R4 g4 r. c+ _7 o$ xThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)! n* k6 ?- R9 k% p% q9 @
= K^3 + 3K^2 + 2K
1 ]$ |& h* V1 z = ( K^3 – K) + ( 3K^2 + 3K)
& r4 d. h$ M+ ^5 {! M+ c5 u; L = ( K^3 – K) + 3 ( K^2 + K): h+ M V& k( O' o" d
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
$ j& g3 M/ U% D9 s8 f3 u9 D& `So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K); l' N4 y& B& S) Q
= 3X + 3 ( K^2 + K)
- Z D1 X+ m/ C: U8 Q b4 @ = 3(X+ K^2 + K) which can be divided by 3
7 t0 t7 @% ?6 ^& e1 }
+ _% U" h" G* u# u8 IConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.) }" Z, X; v7 M' v
( u2 T+ K+ X: I; _+ H% v7 u
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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