The target series is #S = sum_{n=0}^{infty}n^3((x+1)/2)^n# with generic term
#t=n^3((x+1)/2)^n#
Let us introduce #S_b = sum_{n=0}^{infty}((x+1)/2)^n# with generic term #t_b=((x+1)/2)^n#
#S_b# is convergent for #-3 < x < 1#. Comparing terms we have
#D_3=d^3/(dx^3)t_b=(n(n-1)(n-2))/(2^3)((x+1)/2)^{n-3} = #
#n^3/(2^3)((x+1)/2)^{n-3}-(3n^2)/(2^3)((x+1)/2)^{n-3}+(2n)/(2^3)((x+1)/2)^{n-3}#
then
#n^3((x+1)/2)^n = 2^3((x+1)/2)^3d^3/(dx^3)t+3n^2((x+1)/2)^n-2n((x+1)/2)^n#
Also
#D_2=d^2/(dx^2)t_b = (n(n-1))/n((x+1)/2)^{n-2} # and analogously
#2^2((x+1)/2)^2d^2/(dx^2)t_b = n^2((x+1)/2)^n-n((x+1)/2)^n#
#2((x+1)/2)d/(dx)t_b = n((x+1)/2)^n#
#2((x+1)/2)D_1= n((x+1)/2)^n#
Putting all together
#S_0 = sum_{n=0}^{infty}{2^3 ((x + 1)/2)^3 D_3 + 3 xx 2^2 ((x + 1)/2)^2 D_2 + 2 ((x + 1)/2) D_1}#
or
#S_o= (2 (1 + x) (13 + 10 x + x^2))/(x-1)^4# for #-infty < x < 1#
Attached is a figure with the comparison between #S# and
#S_0 =(2 (1 + x) (13 + 10 x + x^2))/(x-1)^4#
Note:
The final expression for #S_0# is obtained considering that #D_k, {k = 1,2,3}# is applied on the equivalence #S_b equiv 1/(1-((x+1)/2)# for #-3 < x < 1#
so
#D_1 = 2/(x-1)^2#
#D_2 = -4/(x-1)^3#
#D_3 = 12/(x-1)^4 #
