求教feifei数学第二部分/第六题

chamcham

1+X+X^2+X^3+X^4+X^5<1/(1-X)?
1)x>0
2)x<1
如何证明c, 当0<x<1时以上等式成立.

rudder86

1+x+x^2+x^3+x^4 = (1-x^5)/(1-x).  
So if 1>x>0, 1+x+x^2+x^3+x^4 < 1/(1-x)
   if x>1, 1+x+x^2+x^3+x^4 >1/(1-x)
   if x = 0, 1+x+x^2+x^3+x^4 =1/（1-x）
if x <0, 1+x+x^2+x^3+x^4 >1/（1-x）?

chamcham

请问次等式如何迩来?
1+x+x^2+x^3+x^4 = (1-x^5)/(1-x).

chamcham

我觉得应该选b

rudder86

LZ, 等比数列求和公式可不能忘了呀！

If x<0, then 1-x^5>1, but 1-x >0.  So (1-x^5)/(1-x) > 1/(1-x)
If x=0, then 1-x^5=1

So when x<1, it is not certain whether 1+X+X^2+X^3+X^4+X^5<1/(1-X)

cckjenius

等比数列 Sn=a1(1-q^n)/(1-q)

1+X+X^2+X^3+X^4+X^5=(1-x^6)/(1-x)

So what we need to know is "(1-x^6)/(1-x)<1/(1-X)?"
1.x≠1
2.If x>1, then 1-x<0 , 1-x^6>1, x^6<0,  no solution
3.If x<1, then 1-x>0, 1-x^6<1, x^6>0, x<1 and x≠0

As a result, x<1 and x≠0

rudder86

Oops.  I was wrong!