Find the sum of series?

Question

(n^2) * (1/3)^n

n=0 to infinity

Please help me out!!

Answer 1

Σ r^n = 1/(1 - r) .. . . n = 0 to ∞

now, differentiating wrt r we have

Σ n r^(n-1) = (1-r)^(-2)

thus

Σ n r^n = r/(1-r)^2 .. . . n is again from 0 to ∞

if we differentiate this again wrt r we have

Σ n^2 r^(n-1) = (1-r)^(-2) + r*(-2)(1-r)^(-3)*(-1)

= [(1-r) + 2r]/(1-r)^3

= (1+r)/(1-r)^3

thus

Σ n^2 r^n = (r + r^2)/(1 - r)^3

and you know the value of r .. . .. .