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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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) T2 H1 W5 W( X$ bProof:
" A) I5 k7 y( D B6 [Let n >1 be an integer
3 f( i0 i' `2 U+ a- h# Z) aBasis: (n=2)+ q0 n5 d' S& @5 M& [# y: a
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 36 w) M8 S) D( J- T; [3 ]9 k
: E5 B ^( V+ e& z. ]; fInduction Hypothesis: Let K >=2 be integers, support that
; b* K k$ A/ u" j3 W d K^3 – K can by divided by 3.
. Z% A$ q) I3 c; L% f4 Y' a9 q" n. \% I6 I$ N. T
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
: b# H5 S; r; O& Q' Qsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
0 d, ]: V& j& H9 Q) P; vThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)+ P- `) d9 X6 o
= K^3 + 3K^2 + 2K8 |1 V6 {2 M4 U0 ?, Q: ^3 j
= ( K^3 – K) + ( 3K^2 + 3K)# T8 T4 R+ o. i8 O8 E
= ( K^3 – K) + 3 ( K^2 + K)
& c; l0 f. B; h* 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>0
0 V8 b' i$ N) {/ j8 |' ~1 wSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
O& m+ r+ N# Y" v+ ` = 3X + 3 ( K^2 + K)
6 \* W, c9 X/ Q: l9 n! m6 o = 3(X+ K^2 + K) which can be divided by 3/ X" w! H5 v" q1 ?
% n% X4 q9 r6 v- }) `8 hConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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