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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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' M) \6 Y w! D, Y* CProof: 1 V( h$ @9 U) d4 [9 {
Let n >1 be an integer
# S+ i) W8 z" _Basis: (n=2)7 k: k2 c' L% R. U/ Z
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
$ Z0 s; f$ c7 P4 I ?1 n5 w) L
5 `( V W4 _6 W$ i: ^Induction Hypothesis: Let K >=2 be integers, support that
) z* c! C) G/ _* c- M K^3 – K can by divided by 3.; {$ ^9 G: E' ^" s
3 V: n- x s! X; U- Y( YNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, _1 B0 R" ?' R7 `1 r5 h" S, usince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem8 u S6 Q4 ~7 D) L
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
" M" K: L L' j a' C = K^3 + 3K^2 + 2K
& J1 J# R* T a, s0 b4 N = ( K^3 – K) + ( 3K^2 + 3K)
" H( Y8 h$ r- X: _. N = ( K^3 – K) + 3 ( K^2 + K); w' D$ C, i* Q' C; k. 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
$ |& e |, W. ESo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
- Q. R, {0 w7 S6 i3 { = 3X + 3 ( K^2 + K)/ T) \3 k; s& D6 x" g! K0 _ F
= 3(X+ K^2 + K) which can be divided by 39 R( k' n. i+ a9 R+ q( T' q$ M
+ R/ \0 o6 q5 v4 @: Y
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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$ O) \3 r$ M: w6 @[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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