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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)8 r s, o% ^9 s
: V$ H1 T7 R2 f/ e, xProof:
0 O9 \4 q' O. W$ |6 q2 `3 tLet n >1 be an integer
/ w' r, e; f2 C9 h. RBasis: (n=2)
* }: O0 i; r( M. r& | 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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% |/ a+ x7 G( u9 wInduction Hypothesis: Let K >=2 be integers, support that8 g2 g/ }. x8 j) d% U w; `! q! Y# Y
K^3 – K can by divided by 3.' s+ G( l5 C ?8 Z
: j8 R& F* a" z* l1 G FNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
( d: F: q7 y2 V: M1 y3 Q3 `& c& wsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem3 M# f; o# f8 S8 a0 v7 V" ^
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)& B7 v: q/ D/ G* T U3 M- O
= K^3 + 3K^2 + 2K
. P- J: q+ H: y/ ]) i; p4 k = ( K^3 – K) + ( 3K^2 + 3K)
! ~. e$ o- [) s, ^" T = ( K^3 – K) + 3 ( K^2 + K)0 v( T( Y. A5 I6 A
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
+ I( ?1 F7 C. S' ?So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
& D5 U, Z' @ N/ ^! T5 @ = 3X + 3 ( K^2 + K)' R, o6 ], S( ~( O
= 3(X+ K^2 + K) which can be divided by 3
( ]$ E+ c0 t$ e, o D% j2 X6 h3 N; c+ d' r/ h0 o( z
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
+ L4 r/ S( C. X s$ O( v# ?3 E# `; {5 {% [
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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