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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); P3 T2 q1 M0 i) u' l
+ \* w b8 G" U& }Proof: * ~% C2 n3 Q% g7 b( o2 I* c
Let n >1 be an integer : I0 K8 [% G, S6 @2 D: V1 [7 v4 ^/ f
Basis: (n=2)
2 [8 `& F( o$ o% P2 v' c 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
) m8 \( x7 B% f7 Q7 G2 z& F) B4 {% x* H' _# ]
Induction Hypothesis: Let K >=2 be integers, support that
3 j6 Q0 V! O& E( j6 S4 W K^3 – K can by divided by 3.' W, l! P% D+ t/ _
3 S' D+ k5 e2 S0 t# f7 y
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3: k% o* l" v, X, S/ X" Q/ G. A
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, ] r7 y! `/ a- T1 j
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1), q7 w5 @ y/ y
= K^3 + 3K^2 + 2K( f( C* e" W. L' } l; l" ?+ U
= ( K^3 – K) + ( 3K^2 + 3K)
! ?* d& X( S/ y7 Q = ( K^3 – K) + 3 ( K^2 + K)
9 d2 s2 [9 U1 Q# @2 a. j; zby 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
4 G0 P; g4 e9 i; NSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)0 `' t; R. r) ^3 L/ J1 }
= 3X + 3 ( K^2 + K)
8 n6 M7 q" {" S! g = 3(X+ K^2 + K) which can be divided by 3
3 x% h( B7 R- U1 u# S' l2 U5 \ O- L q5 D$ J& i0 ]* f
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.# W) g9 k, s, `$ K4 j
: E3 U+ l* D( q4 M8 X
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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