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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)) I2 V8 c5 J& F I
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Proof:
- [; r1 n$ X7 H" l( D" P! fLet n >1 be an integer
+ Q( V6 h$ ^* }Basis: (n=2)
. L1 Y! Y+ c) @; y 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' F! w% ~1 n, |; D! |( M* M# K
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Induction Hypothesis: Let K >=2 be integers, support that' V4 W3 C. N7 G2 o5 e% e( M
K^3 – K can by divided by 3.) R- R$ x7 _; ]* ]& ] r
$ P* t$ u- q$ n j: p3 x! pNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3! _% D2 w2 ?# s1 t1 x" i5 E
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
' }7 \' {! F& ZThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
5 W+ W2 b* W6 d2 ]) c/ x& n = K^3 + 3K^2 + 2K
( _, [. v* ^3 L/ {' _- C1 z4 c0 } = ( K^3 – K) + ( 3K^2 + 3K)
! ^4 ~& K% ~. e = ( K^3 – K) + 3 ( K^2 + K)
' _5 m* j/ ~+ H* n" J/ ]; Zby 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% T6 n5 V' K/ _9 e$ t/ x
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
2 E$ O$ D8 K) t5 Q8 _( q* o- F = 3X + 3 ( K^2 + K)
4 b: d! }) ~# x* z. k3 ? = 3(X+ K^2 + K) which can be divided by 3: J9 T1 J% b" D# j
9 [6 Q a6 z) U# q8 d* S5 sConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.% j, `# S0 T3 }8 o; J8 g
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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