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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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3 h$ p$ J: S# S8 C- C' OProof: O ~. K. `& D
Let n >1 be an integer
% ]6 m3 U. [7 @. m! |" x! W" ]Basis: (n=2)
9 V z7 q0 q/ g# z 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 35 W& _1 f8 V5 l/ j1 T4 _# V$ _5 ^
1 o* h1 B2 U1 [; ^/ \! k
Induction Hypothesis: Let K >=2 be integers, support that
' B2 h; k6 M# e, \' O K^3 – K can by divided by 3.
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& C: A+ k3 ~* w7 @- ?Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3/ ~ O: t9 Y7 t3 Y: C
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
+ G/ X3 Q; B0 {. J) UThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)% J$ v) t( S+ O$ P
= K^3 + 3K^2 + 2K
( e: O5 o9 P! Q+ h8 X3 @3 v5 L = ( K^3 – K) + ( 3K^2 + 3K)) {1 q! ]0 i/ w5 q3 U! [' S- B
= ( K^3 – K) + 3 ( K^2 + K)6 E3 P+ O, L @: a* w* i
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- M5 Q& }' ~) r) t+ ?0 m
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
. W* u6 X; w" V6 _7 `/ I = 3X + 3 ( K^2 + K)( e7 f5 O) a+ ~1 u8 d7 ]( q
= 3(X+ K^2 + K) which can be divided by 3
0 ]; n! v, [- Y4 q- B2 u- @( T" ?2 W; W4 _; T
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.+ D8 L0 L8 ~' ~$ D6 F- S J
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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