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!

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!

Answer 1

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