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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)1 \' ~+ T4 [8 t) ^
7 K- H; _3 R, F, t1 D; dProof: 9 t W! \9 j" e5 w3 t. J
Let n >1 be an integer # a( V: {& J5 D/ h3 B8 F/ w
Basis: (n=2)
. G1 p# `, `+ m0 ^. \$ p& r 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 37 u8 a. l9 m5 J1 e
; w9 L& H1 M9 L; pInduction Hypothesis: Let K >=2 be integers, support that+ u. Q" e0 b4 \& T
K^3 – K can by divided by 3.# s$ X- ^8 {0 e, n) d
/ F6 h- f& g, h' y2 k
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3# h+ P- X/ r' z; f
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem( k- L0 f- h. h6 T4 H' F8 E# y; }
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
: b9 H8 q/ I) D = K^3 + 3K^2 + 2K* k- p5 c9 f! C* K
= ( K^3 – K) + ( 3K^2 + 3K)
1 r. C* ] H. e+ z7 z& o/ F5 ^ = ( K^3 – K) + 3 ( K^2 + K)* w& l4 k3 X# h- }7 {/ 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
8 e; j2 k; f: {& [5 mSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)9 j" O! ^8 m3 b9 f: I7 ^
= 3X + 3 ( K^2 + K)
2 Y4 Y' _% N8 m+ A- d$ E A) Q4 G = 3(X+ K^2 + K) which can be divided by 3: ]! ^' t3 s- w8 D
H, X! N& U. w4 fConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.2 v! j" M, K# _( Z* A7 K
: l8 S6 w6 G6 m d
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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