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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)
* c8 ~- y S& o+ r1 f4 q( {2 K6 U. o; m1 k+ Y, m7 `
Proof:
" F! I1 y" a0 Y# E/ m9 B1 B: rLet n >1 be an integer
2 ?4 C) u$ o7 RBasis: (n=2)# c8 o: G3 B% [" C+ [' D0 I
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3" Z! E/ Q6 N* E z* G
3 q" R; Q" V1 _' Z5 EInduction Hypothesis: Let K >=2 be integers, support that
; ~* l5 C" b. I. |! p0 t K^3 – K can by divided by 3.
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3" n* u: M5 v* t$ Z, G6 ]
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
0 b( [3 c' v4 B3 E) V) cThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
+ U D+ T# Y# K6 Q = K^3 + 3K^2 + 2K# P: Z* u* I W4 `$ ~& n
= ( K^3 – K) + ( 3K^2 + 3K)
/ t1 t5 V6 B; x) U. U = ( K^3 – K) + 3 ( K^2 + K)
# q6 u9 Z: W0 `; Mby 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
: J: F2 Q( {3 k2 J cSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)5 l0 ^% n' x* u" i* @0 k
= 3X + 3 ( K^2 + K)* M7 K6 ~0 p+ F6 @/ X/ z: g
= 3(X+ K^2 + K) which can be divided by 3, w, Z9 f+ P. r+ N: q
: s- I- m% S& |Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.: n" I2 Q3 m; w0 H( N3 a/ x
2 l* I: B" C" x# H% L[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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