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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)0 N/ {# O7 Q# \3 e8 c4 {. a( z! I
$ L2 \9 a5 B$ ?3 u UProof: / l0 S9 V& K/ f( W
Let n >1 be an integer 4 Y- d+ S5 J0 S9 j0 X
Basis: (n=2)
3 T8 W; \% t: f- g: A# N' {8 M 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3" V1 O1 `/ Q& L" Q: }0 ^
5 p! G+ h% J) V4 j* nInduction Hypothesis: Let K >=2 be integers, support that
' ~& s' K. m* a2 d k K^3 – K can by divided by 3." M2 k$ V7 s9 K
9 T+ o! U# h7 `% {
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
3 U+ P( Y1 K( Psince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
2 A, w' b2 F" N! kThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
+ H) i" V9 E8 R. v = K^3 + 3K^2 + 2K
% n) i" Q4 I0 k2 r; b = ( K^3 – K) + ( 3K^2 + 3K)* H N. n: t% u1 n
= ( K^3 – K) + 3 ( K^2 + K)
% t$ l# F) V. Z4 d$ Cby 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% s. ], L, H( L0 R
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
" x( b0 V9 J9 `( x- `5 K' R = 3X + 3 ( K^2 + K)
1 b* U: P) r/ ^8 K+ z0 ~& W = 3(X+ K^2 + K) which can be divided by 3
0 J; K. C+ b. G; l9 ]% C" G" c' k( y8 _7 c& {) z
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
5 H# \1 S) B2 R/ F. `
$ Q; F% i, @) g3 W[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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