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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)+ d3 s) S) @$ y; i
& l) }: @# Y6 w R7 q/ e6 [2 @Proof:
5 _" u2 q- y: q9 t5 ]$ fLet n >1 be an integer
" \9 L/ X5 ]% Y4 W" s% B jBasis: (n=2)
4 o- \7 L& J4 z& n* |1 B- ~! s 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 32 X7 C# G& y$ V( H
7 ` \# U, l7 u+ U5 Z2 A
Induction Hypothesis: Let K >=2 be integers, support that
w* u; B+ P( q# y2 q* X& I K^3 – K can by divided by 3./ N& P. t3 ~5 |" R4 P
' x4 ^, ]0 Z9 ^' g
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
% ?& h1 M# Q* P; [$ O7 _( X ysince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
' ^) j7 F' ?8 q( I- pThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
( x7 y4 W2 W1 u3 u = K^3 + 3K^2 + 2K9 ]- E. f9 ~+ V
= ( K^3 – K) + ( 3K^2 + 3K)! ~4 F4 v2 B' T2 i) s6 t" M
= ( K^3 – K) + 3 ( K^2 + K)/ |) [' k6 Z$ L4 y
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>01 i( f2 x! y: X; c5 P. {0 n
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K), ?: k3 i" r7 ]4 ]+ k( G
= 3X + 3 ( K^2 + K)
5 G: }* Z$ G1 i8 a/ s = 3(X+ K^2 + K) which can be divided by 3: F) F! j$ n! n7 I+ ] o
: d$ K) w( R0 B
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
2 F$ \! H: t0 C0 k8 n* c. q; \8 s. R5 T1 W w" S3 O1 r, I
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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