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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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$ V6 C; G9 d! d% uProof:
+ t2 s' D9 c Y+ `- K# y' mLet n >1 be an integer / R4 }- m9 t9 N$ v# L
Basis: (n=2)" G5 W! n/ b& U. k8 T3 ?. Q! x
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that4 H' m, A) J' k8 i/ U
K^3 – K can by divided by 3.
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) L: _0 [4 @% r% ?; VNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3, B7 t4 g, h7 _: w! y3 W* |+ S
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
, ^$ \6 G2 w8 q+ t0 d. a: ~+ N4 ?Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
; N; c/ H9 E7 s6 ]2 { = K^3 + 3K^2 + 2K; p o" Q/ p1 B- @4 r2 _
= ( K^3 – K) + ( 3K^2 + 3K)) _+ @. ?" v) j5 b
= ( K^3 – K) + 3 ( K^2 + K)
& D2 G9 s7 r) d/ | G; aby 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
% q7 v! o* U% {% {( }0 s5 \; Q* g7 ySo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K); t6 N" _' p% f5 W. Y$ _
= 3X + 3 ( K^2 + K)
6 E9 d8 S9 }* W2 T0 J = 3(X+ K^2 + K) which can be divided by 3' u+ ^& c1 u6 @& B4 a; G! H6 G
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.3 y3 T9 g! o' A7 k) \
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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