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^n-1) - n^2consider1+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)^2put x=2 summation(i:1->n)(i。2^(i-1))= n2^(n+1) -(n+1)2^n +1=(n-1)。2^n + 1Tn = 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