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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)) R' l- W+ K0 ]3 p4 S
& ]' ^+ { }% l0 X$ c/ A5 bProof:
6 h, f8 B1 |$ a$ Q/ S5 e0 I$ }Let n >1 be an integer
3 P. V) e& t0 [8 w4 V9 k5 Z3 zBasis: (n=2)
9 A# L/ l# } I2 Y1 b# v 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3" I$ Y0 P, g. h9 F+ e7 F6 g
% k. n# P2 x! v" a6 ^Induction Hypothesis: Let K >=2 be integers, support that2 h b7 Q7 I7 \( S5 O2 @2 D
K^3 – K can by divided by 3.
6 p" O8 }/ p2 a& p
$ `5 ~2 V3 e7 u. L3 |Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
% Z) N# e7 \3 B0 i' gsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem9 D# A ^5 `9 _" N3 X) x
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' m/ ^* Z& J" [5 n; k l2 @, E
= K^3 + 3K^2 + 2K6 L5 K. ?5 u$ _, h2 K2 q
= ( K^3 – K) + ( 3K^2 + 3K)9 h- j6 C ^9 u% @' y
= ( K^3 – K) + 3 ( K^2 + K)8 ?5 S) `: ?1 Q# K1 l' a% j
by Induction Hypothesis, we know that k^3 – k can by divided by 3 which means that k^3 – k = 3 X for some integer X>05 o4 t5 \/ d+ K Z$ n0 e# m5 \) j
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)% F! O: p# ]" C3 t. z( ~- g
= 3X + 3 ( K^2 + K)
" ~. L0 u, P4 q* x = 3(X+ K^2 + K) which can be divided by 3/ U: Z6 J1 u8 I/ l+ G5 B
& d5 `9 v2 r2 S$ J7 NConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.3 i1 A- o* t# U2 R% y
3 l/ E2 s8 Y y3 k5 U3 W[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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