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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)( j+ }$ x. E) u# a
1 H y- f1 Y$ n1 PProof:
' z* \, q4 V i# ~, i( r/ oLet n >1 be an integer ; U8 w7 E5 w% [$ j
Basis: (n=2)
( s: `5 M: x. y1 _* v. O: M+ z 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3& H; K+ m5 J# G
! r1 ]; T7 \. B2 M
Induction Hypothesis: Let K >=2 be integers, support that
+ Y2 i& y9 U3 r) Z0 w K^3 – K can by divided by 3.: a8 T: J r! Y3 G
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3* A, L% U- P* F' x% u0 i
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
1 G+ `& Q5 q+ n9 r4 o F* ^. ?7 `: IThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
1 V, U! G2 J; d( U0 A9 e = K^3 + 3K^2 + 2K
+ G1 j) q+ ]$ w0 m' |/ H2 | = ( K^3 – K) + ( 3K^2 + 3K)
- A X+ h }/ x" o* } = ( K^3 – K) + 3 ( K^2 + K)/ g1 q! ]- B7 }: q% o
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>00 I0 z5 I% b! m5 s/ s7 g) X
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ y9 N. H' u3 ~ = 3X + 3 ( K^2 + K)4 ~5 G4 q0 w$ t, x4 w, V0 I
= 3(X+ K^2 + K) which can be divided by 3! E+ I# L0 W. @$ W
/ U0 s! D' {% v' U( k) OConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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! D8 ~; m7 H% K9 g9 p[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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