 鲜花( 53)  鸡蛋( 0)
|
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 Y5 L' l5 `6 I* I9 c) k0 V( v2 z4 V, Y2 t# }
Proof: . S* Z9 h9 b7 c# I
Let n >1 be an integer
+ g& h* n- I! |/ r8 Z$ kBasis: (n=2): N: u+ q1 J0 U4 s6 l* [
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 32 E" o" U6 s" o8 c
& G8 W6 y/ b5 j6 G" s1 X
Induction Hypothesis: Let K >=2 be integers, support that9 }9 Z; F* v7 ^& ^
K^3 – K can by divided by 3.$ T' f5 ~+ `3 p, y
# z/ B* d& H# V. mNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 35 c# }& a& s$ _' V! \& y
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
+ w- K! o# R: j' L/ v: g: a$ JThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
% |2 c5 Q) X+ ~' F" t) N) i& g = K^3 + 3K^2 + 2K7 Y) w \' ~ o# T* t
= ( K^3 – K) + ( 3K^2 + 3K)
! l+ P4 l3 A, B* Z8 y( F = ( K^3 – K) + 3 ( K^2 + K)% V* M1 ` ]( v- f( ^. M' E9 h
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>02 M8 B0 {% K% e- }) e, a P
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)/ ]2 I! t% \2 P3 `- U# m
= 3X + 3 ( K^2 + K)
" o7 G8 x; J7 S = 3(X+ K^2 + K) which can be divided by 3
4 N4 t5 l- h+ T/ Q' P$ @2 V
# s6 g8 I2 P7 S! i0 pConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
: Y. A& a2 Y, W' C# {, p3 B- \' p4 \) O2 W; L' C
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|