设数列｛an｝的前n项和为Sn,且 Sn=(x+1)-xan,其中x是不等于-1和0的常数.①证明｛an｝是等比数列；②设设数列｛_数学解答

设数列｛an｝的前n项和为Sn,且 Sn=(x+1)-xan,其中x是不等于-1和0的常数.①证明｛an｝是等比数列；②设
设数列｛an｝的前n项和为Sn,且 Sn=(x+1)-xan,其中x是不等于-1和0的常数.
①证明｛an｝是等比数列；②设数列｛an｝的公比q=f(x),数列｛bn｝满足b1=1/3,bn=f{b(n-1)}(n 属于N,n>=2),求数列｛1/bn｝的前n项和Tn.
求详解第二问,第一问已解.

人气：281 ℃　时间：2020-03-26 04:39:48

解答

(1)
Sn=(x+1)-xan (1)
S(n-1)=(x+1) -xa(n-1) (2)
(1) -(2)
an = xan-xan(n-1)
an/a(n-1) = x/(x-1)
=>{an}是等比数列
(2)
q=f(x) =x/(x-1)
bn=f(b(n-1))
= b(n-1)/[b(n-1) -1]
1/bn = [b(n-1) -1]/b(n-1)
= 1- 1/b(n-1)
1/bn -1/2 = -(1/b(n-1) -1/2 )
= (-1)^(n-1) .( 1/b1- 1/2)
= (5/2) .(-1)^(n-1)
1/bn = 1/2 +(5/2) .(-1)^(n-1)
Tn = 1/b1+1/b2+...+1/bn
= n/2 + (5/4)( 1- (-1)^n )第一问算错正负号了吧。。。Sn=(x+1)-xan (1)S(n-1)=(x+1) -xa(n-1) (2)(1) -(2)an = -xan+xan(n-1)an/a(n-1) = x/(1+x)=>{an}是等比数列(2)q=f(x) =x/(1+x)bn=f(b(n-1)) = b(n-1)/[1+b(n-1)]1/bn = [1+b(n-1) ]/b(n-1) = 1+ 1/b(n-1)1/bn -1/b(n-1)= 11/bn -1/b1 = n-1 = n+2Tn = 3+4+...+(n+2) = (n+5)n/2