an=2的n-1次方,bn=2n-1,设数列an的前n项和为sn,求数列{sn.bn}的前n项和Tn

an=2^(n-1)bn=2n-1Sn = 2^n -1Sn.bn = (2^n -1)(2n-1)= 4.(n2^(n-1)) - 2^n - 2n +1Tn = S1b1+S2b2+...+Snbn= 4{ summation(i:1->n)(i.2^(i-1)) } - 2(2^n-1) - n(n+1) +n= 4{ summation(i:1->n)(i.2^(i-1)) } - 2(2...