[tex]\displaystyle a_n =\sum\limits_{k=1}^{n}\dfrac{k(k+1)}{2x^{k-1}} \\ \\\dfrac{1}{x} = t,\, |x|>1\Rightarrow |t| < 1 \\ \\ a_n = \sum\limits_{k=1}^{n}\dfrac{k(k+1)}{2}t^{k-1} \\ \\ \\|t|< 1\\ \\ \dfrac{1}{1-t} = 1+t+t^2+t^3+...\Big|' \\ \\ \dfrac{1}{(1-t)^2} = 1+2t+3t^2+4t^3+...\Big|' \\ \\ \dfrac{2}{(1-t)^3} = 1\cdot 2+2\cdot 3t+3\cdot 4t^2+4\cdot 5t^3+... \Big|:2[/tex]
[tex]\displaystyle \dfrac{1}{(1-t)^3} = \dfrac{1\cdot 2+2\cdot 3t+3\cdot 4t^2+4\cdot 5t^3+...}{2} \\ \\ \dfrac{1}{(1-t)^3} =\sum\limits_{k=1}^{\infty}\dfrac{k(k+1)}{2}t^{k-1} \\ \\ \\\text{Revenim la totatie:}\\ \\ \dfrac{1}{\Big(1-\dfrac{1}{x}\Big)^3} = \sum\limits_{k=1}^{\infty}\dfrac{k(k+1)}{2x^{k-1}} \\ \\ \\ \Rightarrow \boxed{\sum\limits_{k=1}^{\infty}\dfrac{k(k+1)}{2x^{k-1}} = \dfrac{x^3}{(x-1)^3}}[/tex]
