数列求和问题an=n^2bn=q^n求an*bn的前n项和

Question

数列求和问题
an=n^2 bn=q^n
求an*bn的前n项和

Answer 1

let
S1 = 1.q^1+2.q^2+.n.q^n (1)
qS1 = 1.q^2+2.q^3+.n.q^(n+1) (2)
(1)-(2)
(1-q)S1 = (q+q^2+...+q^n)-n.q^(n+1)
= q(1-q^n)/(1-q) -n.q^(n+1)
S1 =q(1-q^n)/(1-q)^2 -n.q^(n+1)/(1-q)
let
S2 = (1.2)q +(2.3)q^2+.+n(n+1)q^n (3)
qS2 = (1.2)q^2 +(2.3)q^3+.+n(n+1)q^(n+1) (4)
(3) -(4)
(1-q)S2 = 2q + 4q +6q^2+.+2nq^n - n(n+1)q^(n+1)
=2(q+2q^2+...+nq^n) - n(n+1)q^(n+1)
=2S1 - n(n+1)q^(n+1)
=2[q(1-q^n)/(1-q)^2 -n.q^(n+1)/(1-q) ] -n(n+1)q^(n+1)
S2 = 2[q(1-q^n)/(1-q)^3 -n.q^(n+1)/(1-q)^2 ] -n(n+1)q^(n+1)/(1-q)
cn = an.bn
=n^2.q^n
= [n(n+1) - n].q^n
= n(n+1).q^n - n.q^n
Tn=c1+c2+...+cn
= S2 -S1
= 2[q(1-q^n)/(1-q)^3 -n.q^(n+1)/(1-q)^2 ] -n(n+1)q^(n+1)/(1-q)
-[q(1-q^n)/(1-q)^2 -n.q^(n+1)/(1-q) ]
= 2[q(1-q^n)/(1-q)^3 -n.q^(n+1)/(1-q)^2 ] -n^2q^(n+1)/(1-q) -q(1-q^n)/(1-q)^2