This sequence has generating function
[tex]F(x)=\displaystyle\sum_{k\ge0}k^3x^k[/tex]
(if we include [tex]k=0[/tex] for a moment)
Recall that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k[/tex]
Take the derivative to get
[tex]\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k[/tex]
[tex]\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k[/tex]
Take the derivative again:
[tex]\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k[/tex]
[tex]\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k[/tex]
Take the derivative one more time:
[tex]\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k[/tex]
[tex]\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k[/tex]
so we have
[tex]\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}[/tex]
