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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)# X' G W7 c b" i& q2 N+ j
, E" B6 j: D9 W( }9 {$ c5 W- C/ GProof:
0 [2 _6 ^+ g7 s9 I5 mLet n >1 be an integer 0 }% b4 V1 p) p' s2 ?; c: J
Basis: (n=2)
6 B9 b( d+ Q0 m5 h& I* P 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
- ?; Y8 Z& P8 X0 Q& ]4 @) G0 m, Z7 V2 l
Induction Hypothesis: Let K >=2 be integers, support that
7 y h4 w. T" R" V$ Z; w- b K^3 – K can by divided by 3.9 r9 S! ?# t( z- w
' L" H. M& n( F+ Y+ Q) iNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
3 g- A; i4 X3 N/ w; esince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
+ Y4 L6 |/ N) i; _$ _( RThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)8 M2 B) a1 Y, @" y* i
= K^3 + 3K^2 + 2K
* s; J" b5 V+ w2 { i; u = ( K^3 – K) + ( 3K^2 + 3K)
! j; k9 C+ A+ X0 N = ( K^3 – K) + 3 ( K^2 + K)8 { O h( |1 j. U
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, A# h, i. k8 s
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
6 m! K2 b) T9 B5 z0 \1 C = 3X + 3 ( K^2 + K): V4 L4 n( r5 D2 d! }
= 3(X+ K^2 + K) which can be divided by 30 {" U1 a& e" b
% _4 Q" d& F+ l: k; T
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
6 h9 n6 y2 Q4 R& }4 {
2 l! D! Z* P- H9 n[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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