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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)7 z" P0 Z9 l! \+ j
% Y& S5 H% A6 ZProof:
. o9 Q- T1 x( `" yLet n >1 be an integer
3 e! l: v: Z' d% mBasis: (n=2)
4 L3 Q- {9 \& `- F; R! R# T 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3: ?* I! z1 k* I. D! h+ i7 p
* B8 \+ e7 O, ?/ RInduction Hypothesis: Let K >=2 be integers, support that' s/ }! C4 F+ J% E& a2 Q- p+ @4 D
K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3$ k" e/ f+ z0 }7 G, |7 b4 {$ |
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
0 r. q$ A; `- x" }# i2 D0 H8 wThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
$ m( q" S3 D6 Q5 t = K^3 + 3K^2 + 2K0 G1 R! ^. M$ R' C( w# E
= ( K^3 – K) + ( 3K^2 + 3K). U; z, |3 T6 j8 i& j
= ( K^3 – K) + 3 ( K^2 + K)7 v5 u8 ]+ E& H: H ?
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 q9 u4 C6 e. U0 R. D) n- S9 ]1 ]
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K). w; \6 J9 \4 u* C8 N, B2 T
= 3X + 3 ( K^2 + K)/ U8 Z9 j* |( V# S6 h
= 3(X+ K^2 + K) which can be divided by 3, r2 C I1 c1 a! V+ @9 E
" O$ v: D! v# e2 u
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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