设Sn是正项数列an的前n项和,知4Sn＝an^2＋2an－3,求an.知bn＝2^n,求Tn=a1b1+a2a2……+anbn的值

Question

设Sn是正项数列an的前n项和,知4Sn＝an^2＋2an－3,求an.知bn＝2^n,求Tn=a1b1+a2a2……+anbn的值

Answer 1

4s(n)=[a(n)]^2 + 2a(n)-3,
4a(1)=4s(1)=[a(1)]^2 + 2a(1)-3, 0 = [a(1)]^2-2a(1)-3=[a(1)-3][a(1)+1], a(1)=3.
4s(n+1)=[a(n+1)]^2 + 2a(n+1)-3,
4a(n+1)=4s(n+1)-4s(n)=[a(n+1)]^2 + 2a(n+1)-[a(n)]^2 - 2a(n),
0 = [a(n+1)]^2 - 2a(n+1) - [a(n)]^2 - 2a(n)
= [a(n+1)+a(n)][a(n+1)-a(n)] - 2[a(n+1)+a(n)]
=[a(n+1)+a(n)[a(n+1)-a(n)-2],
0 = a(n+1)-a(n)-2,
a(n+1)=a(n)+2,
{a(n)}是首项为a(1)=3,公差为2的等差数列.
a(n)=3+2(n-1)=2n+1.
c(n)=a(n)2^n=(2n+1)2^n,
t(n)=c(1)+c(2)+...+c(n-1)+c(n)=(2*1+1)2+(2*2+1)2^2+...+[2(n-1)+1]2^(n-1)+(2n+1)2^n,
2t(n)=(2*1+1)2^2+(2*2+1)2^3+...+[2(n-1)+1]2^n + (2n+1)2^(n+1),
t(n)=2t(n)-t(n)=-(2*1+1)2-2*2^2-...-2*2^n+(2n+1)2^(n+1)
=-6-2^3(1+2+..+2^(n-2)) + (2n+1)2^(n+1)
=(2n+1)2^(n+1)-6-8[2^(n-1)-1]/(2-1)
=(2n+1)2^(n+1)-6-2^(n+2) + 8
=(2n-1)2^(n+1)+2