已知数列{an}的前n项和为Sn,且Sn=2an-2n n∈N*）（1）求数列{an}的通项公式；

Question

（2）求数列{n.an}前n项和；（3）若数列{bn}满足了bn=1/[log2(an+2)].1/[log2(an+1)+2],求数列{bn}前n项的和Bn

Answer 1

(1)
Sn=2an-2n (1)
n=1, a1=2
S(n-1)=2a(n-1)-2(n-1) (2)
(1)-(2)
an =2an-2a(n-1) -2
an = 2a(n-1) +2
an+2 = 2(a(n-1) +2)
(an+2)/(a(n-1) +2) =2
(an+2)/(a1+2)=2^(n-1)
(an+2) = 2^(n+1)
an = 2^(n+1) -2
(2)
cn = n.an
= n(2^(n+1) -2)
= 4(n.2^(n-1) ) - 2n
Tn =c1+c2+..+cn
= [summation(i:1->n) {4(i.2^(i-1) )} ] - n(n+1)
consider
1+x+x^2+..+x^n = (x^(n+1)-1)/(x-1)
1+2x+..+nx^(n-1)= [(x^(n+1)-1)/(x-1)]'
=[nx^(n+1)-(n+1)x^n +1]/(x-1)^2
put x=2
1.2^0+2.2^1+...+n.2^(n-1)
=[n.2^(n+1)-(n+1)2^n +1]
= 1+ (n-1)2^n
Tn=[summation(i:1->n) {4(i.2^(i-1) )} ] - n(n+1)
= 4[1+ (n-1)2^n] - n(n+1)
(3)
an = 2^(n+1) -2
bn=(1/log<2>(an+2) ) [1/ log<2>(a(n+1)+2 ) ]
=1/[(n+1)(n+2)]
=1/(n+1) -1/(n+2)
Bn =b1+b2+..+bn
= 1/2 -1/(n+2)