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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)
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Proof:
/ @6 i2 I$ c" E# z9 X. b( ^( u% HLet n >1 be an integer
$ C8 R, x) T8 u9 V# [& CBasis: (n=2)
, C3 G F7 E2 Q; Q2 B8 s$ w, q 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 30 L( X8 Y! }; O
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Induction Hypothesis: Let K >=2 be integers, support that" E" o6 H6 `5 L5 @4 A7 y
K^3 – K can by divided by 3.7 H" k; l2 j# E$ @) m. F2 l# h; O
# x3 B. D% W0 V3 @8 \- Q. ZNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3; q. G$ S4 g8 _) H
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
0 O8 ]7 p2 {0 mThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
. A; X ^" B1 Z1 y. a = K^3 + 3K^2 + 2K$ W: w, Q+ J' m9 R% C, r
= ( K^3 – K) + ( 3K^2 + 3K): c' \; `3 B: m( g* K8 D! f
= ( K^3 – K) + 3 ( K^2 + K)
$ |6 t' q( | Y. kby 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
- a) v, n7 E1 }So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)9 j" W6 j; |1 m
= 3X + 3 ( K^2 + K)
7 x& b# f/ \: b8 E = 3(X+ K^2 + K) which can be divided by 36 v4 V7 D- a4 c/ @4 u X- P, S
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1., p" b$ S8 ]$ l/ p: g
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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