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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)8 a' t, e% h- @
6 V+ \+ ^6 B B6 S( xProof:
4 O7 s4 y4 q! g: P I A* @Let n >1 be an integer
/ I) @2 @2 d2 k ~% h; G% gBasis: (n=2)
' Y, u9 `# i1 P7 |3 l 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 33 O' x3 R3 d5 {" n
& E% ^( b; m2 k9 a. K+ W
Induction Hypothesis: Let K >=2 be integers, support that
4 |/ E7 R+ l) c# ]2 O K^3 – K can by divided by 3. L/ ~+ b7 i4 f& h, X
1 ]; K) Q) m$ n1 w' [8 c6 |# h* l
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
0 p& A+ e) N/ w1 j6 d& Q0 dsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem- X, V$ w$ J: p
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)0 n* {* Z( }7 O; W7 q, h
= K^3 + 3K^2 + 2K
% O5 s4 `0 r7 z4 k' h; ~( |5 i = ( K^3 – K) + ( 3K^2 + 3K)
4 w) m2 C- t+ Q+ j! C. B = ( K^3 – K) + 3 ( K^2 + K)' w2 l# C4 p) o
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/ Y5 j' {6 D3 W1 M( b: _" K3 Q1 Y
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)' H# n5 y* |+ A8 T$ ~7 \: C
= 3X + 3 ( K^2 + K)% C+ M) G& |& C$ q
= 3(X+ K^2 + K) which can be divided by 3
% H% c% ^# C% a/ N. M9 @ K+ ]4 I9 y
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
; Z$ q1 {6 L7 E( s: ]+ t5 u& N* E7 V: r, h2 c
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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