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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) I' m& c! m* S( y
& q% t0 t6 P. f0 v* a' D- PProof: 8 Y& m$ X; j! d! ]* V/ O3 d
Let n >1 be an integer
( F6 {+ i+ J4 _7 W* o5 Y& z5 M( f; qBasis: (n=2)+ U( }/ J4 q# @* w- s9 M
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 39 s$ Z2 O1 s% T9 q
) Z" k: F6 ~' \& G5 @
Induction Hypothesis: Let K >=2 be integers, support that
$ ~* p9 g4 e, e4 l& m0 L1 R K^3 – K can by divided by 3.* W3 q/ L1 u( `' v+ D
$ Z9 b2 q( {. l8 bNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3! J. V1 l# A; D; Y
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem/ M2 n, x0 V6 L3 w7 Q5 b% I; ]
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
( U0 G5 J& A$ x k7 d' `: @ = K^3 + 3K^2 + 2K
( v/ f( h# R2 N+ H: _8 Q) f, ]5 o = ( K^3 – K) + ( 3K^2 + 3K)& {; B5 C) y' Q4 ]
= ( K^3 – K) + 3 ( K^2 + K)# R6 K$ g& V5 L \( e' ~- X
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>00 x I4 P+ x4 g3 T
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
1 X( D. d$ @* }/ k6 c. o = 3X + 3 ( K^2 + K)
) b$ C; u- ]& `. r0 O' `9 y; ` = 3(X+ K^2 + K) which can be divided by 39 u% j& y4 N7 Y! [; \
& Z. L# m) | G3 v9 c& m* ^Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.* V' i9 [% Q" T. W1 p
9 J' ~ n1 J! D' N5 f$ F
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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