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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)2 U# y( B! a9 ^* {
& ]! M5 J# ]" r% EProof:
! o9 H8 |0 q) s* I- o" ~3 zLet n >1 be an integer # q6 q0 z3 k3 v. X0 f
Basis: (n=2)
; G* @7 R" P2 @8 Q 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
0 l1 z2 c, ^% @; K J3 b+ r, i
( y# a" s7 {: ^% wInduction Hypothesis: Let K >=2 be integers, support that
" u0 K3 f& |$ Z6 v$ Z5 } K^3 – K can by divided by 3.
- N }' A+ e7 Y* o/ N- J* ?: K. K- a
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
p5 v6 W* @+ L2 ^. A t4 v. |since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
; T8 w. p1 h6 y7 q2 eThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)5 `" o1 N* }0 i# C+ G6 `9 ~& R
= K^3 + 3K^2 + 2K
7 B4 Z0 s8 S+ J% `$ q; {& M$ x = ( K^3 – K) + ( 3K^2 + 3K)
S ~0 c' v+ O6 i! E! Q& X = ( K^3 – K) + 3 ( K^2 + K)6 o) e8 Y* h" H6 D
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. [/ S( F, V0 O* s9 T+ ]
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)! Y# q* t) {" A4 D: _& A
= 3X + 3 ( K^2 + K)
. T0 D5 g* {" A ^# s: L = 3(X+ K^2 + K) which can be divided by 3
% ^* w3 q! k' J
: a6 s( n; u' G( N* WConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
. b) X" R/ g# ]# J% L4 k# X) X
! L: @3 ~. s. C) A1 P" y7 ?, Z' m/ {[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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