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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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1 A' x" P: M9 k, |$ KProof:
# L2 {. {6 |! TLet n >1 be an integer 9 y% ^( \" M. e: [
Basis: (n=2)( d- M" D& |1 C+ Z% j
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3' y) {3 u0 c% p! g" I) M" }
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Induction Hypothesis: Let K >=2 be integers, support that
3 z3 b( C. F! y1 V% Z" Y6 D6 n& N. o K^3 – K can by divided by 3.# c+ W( q# B# `3 s: r) b
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 ~- D0 c) M* S# T2 L2 U' B# J3 ^since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem3 j& q2 \; q/ ]0 R) T
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)! S# U9 ^$ [! Q
= K^3 + 3K^2 + 2K, |5 g4 H& A! Y Y
= ( K^3 – K) + ( 3K^2 + 3K)5 O! }- @; C* _# O4 O' y, V
= ( K^3 – K) + 3 ( K^2 + K)
+ x7 o! q {1 {" R6 R& s& g3 p$ iby 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, Y1 L9 l. G3 A! W
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
; k" E+ V+ L8 y( A6 k = 3X + 3 ( K^2 + K), z9 q8 W7 g9 R0 a
= 3(X+ K^2 + K) which can be divided by 3
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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/ }- N7 q. J2 e% R[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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