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-1
Sn = 2^n -1
Sn.bn = (2^n -1)(2n-1)
= 4.(n2^(n-1)) - 2^n - 2n +1
Tn = 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^n-1) - n^2
consider
1+x+x^2+..+x^n = (x^(n+1)-1)/(x-1)
1+2x+3x^2+...+nx^(n-1) = [(x^(n+1)-1)/(x-1)]'
= [nx^(n+1) -(n+1)x^n +1]/(x-1)^2
put x=2
summation(i:1->n)(i.2^(i-1))
= n2^(n+1) -(n+1)2^n +1
=(n-1).2^n + 1
Tn = 4{ summation(i:1->n)(i.2^(i-1)) } - 2(2^n-1) - n^2
= 4[(n-1).2^n + 1] - 2(2^n-1) - n^2
= 2(2n-3).(2^n) -n^2 +6