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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)+ K j4 q. Y& C$ i O& D% ]4 k1 F3 F
0 `+ O: K& ?* s8 f
Proof:
% |1 x* x* r5 @, D* |/ N/ TLet n >1 be an integer
$ q% X X3 R |8 T H: {* dBasis: (n=2)
! N* i& ]& W( i: ~" b" } 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 38 X! u* `2 B8 U* T6 G; S* D/ L
! X- V$ p \+ z+ E d: n
Induction Hypothesis: Let K >=2 be integers, support that
4 P. o( Z& g U1 p; F" O K^3 – K can by divided by 3.. [1 |6 X: U6 o( _" H' U6 @- T$ Y
& g* @2 M6 U$ N1 ^Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 32 X% D" A; q. D; n/ @6 i2 N/ F
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
9 }. n5 L7 F2 J9 P$ HThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
, m) _$ k' g$ m# |) s- [8 F = K^3 + 3K^2 + 2K, ]9 a- ?4 q) C4 z) s0 c3 S
= ( K^3 – K) + ( 3K^2 + 3K)4 w; o+ N3 J* j2 y6 v2 K, i. c
= ( K^3 – K) + 3 ( K^2 + K)2 j( Z l+ @8 N9 t4 ]
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. }5 t0 w! l9 n6 ]9 L% f2 x D9 A
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
* [" c/ Z7 x+ O; T% Q( L$ k1 t/ C4 L = 3X + 3 ( K^2 + K)6 f5 U, n: t+ w, Z" Z9 } d9 T* C2 x- ^
= 3(X+ K^2 + K) which can be divided by 3) X. u2 |4 S$ N: n; Z/ {
3 P+ f( S9 i* E" pConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
$ x- [% |( a8 v' ~; a) P- \+ m
* _9 T& x8 o4 t3 _0 }3 `) b[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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