Question

高中数学 数列求解

要带过程 谢谢

Answer 1

(1)数列{an}的前n项和为Sn,且Sn=1−an，

可得Sn−1=1−an−1,两式相减可得：2an=an−1,所以数列{an}是等比数列公比为：12,S1=1−a1，

首项为：12，

an=12n，

(2)bn=log4a1+log4a2+…+log4an=log4(a1a2…+an)=log4(12)1+2+3+…+n=−n(n+1)4

数列{1an+1bn}的通项公式为：2n−4n(n+1)，


数列{1an+1bn}的前n项和Tn=(2+22+23+…+2n)−4(1−12+12−13+…+1n−1n+1)=2(1−2n)1−2−4(1−1n+1)=2n+1+4n+1−6，

Answer 2

(1)
2Sn +1=4an
Sn =(1/2)(4an -1)
n=1, a1= 1/2
for n>=2
an = Sn -S(n-1)
=2an - 2a(n-1)
an =2a(n-1)
an = 2^(n-1) . a1
=2^(n-2)
(2)

b1=2
a(n+1).bn =2an.b(n+1) -2^n
2^(n-1).bn = 2^(n-1).b(n+1) - 2^n
bn = b(n+1) - 2
b(n+1) -bn = 2
bn = b1+2(n-1)
=2+2(n-1)
=2n
cn =log<2>(4an) = n
dn = 1/[b(n+1).cn]
=1/[2n(n+1)]
= (1/2)[ 1/n -1/(n+1) ]
Tn
=d1+d2+...+dn
=(1/2)[ 1- 1/(n+1) ]
=n/[2(n+1)]