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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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4 k' `# B3 P/ x7 q! ^" V5 eProof:
O i& y% Z- M' o$ Y4 u' s' Z: qLet n >1 be an integer
/ \ H) B* i9 x& k; X4 T! M( i% W) cBasis: (n=2)* X4 u$ U1 L2 R+ F' z& O
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 39 Q. O `# |+ Y& o. }4 X
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Induction Hypothesis: Let K >=2 be integers, support that
5 B% _! a5 t4 u6 M K^3 – K can by divided by 3.$ n/ j* I7 z; G) P/ ]6 @3 E
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
: X! A6 l3 U- l v# Asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
7 Z6 w4 V5 W; R$ M# o2 [ `$ o* FThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1), t4 L# E4 z+ q- M3 C3 |8 W
= K^3 + 3K^2 + 2K
2 P8 u) C% @% K0 q' o- K = ( K^3 – K) + ( 3K^2 + 3K)) Z3 g3 r# \2 H! L3 E
= ( K^3 – K) + 3 ( K^2 + K)8 K5 {# g. U. g3 D
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; `6 B H1 b. R( r
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 J' L! K# F* J5 k3 j z9 {' j = 3X + 3 ( K^2 + K)$ ]" E0 I' \* L* r Y
= 3(X+ K^2 + K) which can be divided by 3, h' E0 @; I# d$ u. O
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.( O7 _) a* e+ j7 F) z) E2 g* P! k
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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