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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); m* l) P4 p' n1 H7 O k6 K
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Proof:
+ n _# m s& f- ^+ Y- r1 LLet n >1 be an integer
* p) U5 R9 u; r) d/ z- K* @Basis: (n=2)
. {1 N$ `6 V2 `( [/ Z8 D 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3" A- ~8 V2 c0 W$ M: i8 m/ a
: H( y$ `% B6 {9 xInduction Hypothesis: Let K >=2 be integers, support that
1 ?9 O6 C) I9 R K^3 – K can by divided by 3.7 Y6 C% E3 W8 i, k! e
( ~& X! {. u; H6 y" {( H/ N- F+ j) SNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
9 S8 R% Z7 ^1 Y* M' ^8 _since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
! ^6 g( y$ l b3 U% E4 OThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)3 `- C/ }$ J0 U
= K^3 + 3K^2 + 2K9 k# j2 K v( X1 ]# X8 Q! i
= ( K^3 – K) + ( 3K^2 + 3K)0 C3 O. t5 T2 H) b3 _
= ( K^3 – K) + 3 ( K^2 + K)
# c n- P# i& ]4 x+ zby 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 c, {; |5 s x! v* U
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)+ |/ m2 a0 a/ i. T: a
= 3X + 3 ( K^2 + K)0 X4 u# r' H; E' |# S
= 3(X+ K^2 + K) which can be divided by 3* r# C* J. k" e4 p4 [
9 i U, c* y! g5 \+ j9 h& s7 DConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.' M9 L% @+ Z& T
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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