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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), U- T- s2 O* z, V. S Z/ w) S
' t5 e, |- t& r5 IProof:
. S+ r) `9 C6 aLet n >1 be an integer
1 x1 P( ~$ F _1 LBasis: (n=2)
/ ]! F/ v n' x5 \) I5 y' H$ K 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
5 r1 t7 B; X9 L3 \2 s F4 Q8 ?
# [. R5 |. @* Q& P. J% F0 aInduction Hypothesis: Let K >=2 be integers, support that
% Y2 y9 [% @7 x& Y K^3 – K can by divided by 3.
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- q$ O) | l* \8 {: qNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3) ?) b1 e' c3 n. b& i5 x7 t4 I a
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
1 w3 E, o' ^- E2 ^$ ]Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)8 T1 Z4 K$ i' ~5 d
= K^3 + 3K^2 + 2K! W: @, U' X% M4 S
= ( K^3 – K) + ( 3K^2 + 3K)# ^, `" D" F& T6 }
= ( K^3 – K) + 3 ( K^2 + K)
9 j c: f: c0 i. Pby 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- v0 R4 s% L; b4 M# ~7 i" x, {0 e
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)# ]' o. [. I7 N0 I c. S! {
= 3X + 3 ( K^2 + K)# X* x8 {1 H2 ^1 e0 Z
= 3(X+ K^2 + K) which can be divided by 3
8 I Y( K0 j6 q/ Q! M1 U6 b* D% f u( p8 d
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.2 z; Y& \: B4 E$ m
3 v. ^* T5 E& q
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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