Question

设{an}是各项都为正数的等比数列，其前n项和为sn，且对任意的n属于正整数都有sn=2an-1.

（1）求数列{an}的通项公式。
（2）求数列{nsn}的前n项和Tn。
急急急急急急！

Answer 1

(1)
Sn = 2an-1
n=1, a1=1

an = Sn - S(n-1)
= 2an - 2a(n-1)
an = 2a(n-1)
= 2^(n-1) . a1
=2^(n-1)
(2)
Sn = 2^n -1
nSn = n.2^n -n
Tn = S - n(n+1)/2
S = 1.2^1+2.2^2+.....+n.2^n (1)
2S = 1.2^2+2.2^3+.....+n.2^(n+1) (2)
(2)-(1)
S= n.2^(n+1) - ( 2+2^2+...+2^n)
=n.2^(n+1) -2(2^n -1)
= 2+ (2n-2).2^n
Tn = S - n(n+1)/2
=2+ (2n-2).2^n - n(n+1)/2