xSn - Sn
= Sn(x-1) = [(x^n+1) n^2 ] - [sum n= 1 to n of [(2n-1)x^n]]
so that
Sn = [1/(x-1)]( [(x^n+1) n^2 ] - [sum n= 1 to n of [(2n-1)x^n]])

I must confess I don't know my sums that well, however this site
http://www.shu.edu/projects/reals/numser/t_geom.html
sum from n=zero to inf of a^n = 1/(1-a)
so sum from n=1 to infinity of a^n would be one less or a/(1-a)

also, for sum n=0 to k of nk^n,
http://mathworld.wolfram.com/PowerSum.html
gives the formula:
[x - (n+1)x^(n+1) + nx^(n+1)]/(x-1)^2
we are actually not interested in the n=0 term, but that won't make a difference, as when k = 0, there is no product.

[1/(x-1)][ (x^(n+1) n^2 - 2[sum 1 to inf nx^n] + sum 1 to inf x^n

so our sequence will be:

[1/(x-1)][ (x^(n+1) n^2 - 2[[x - (n+1)x^n+1 + nx^(n+1)]/(x-1)^2] + x/(1-x)]

now, when n approaches infinity, as |x|<1, n^2 . x^n will disappear, and our above expression will become
[1/(x-1)][ - 2[x/(x-1)^2 ] + [x/(1-x)]
[1/(x-1)][ - 2[x/(x-1)^2 ] - [x/(x-1)]
[1/(x-1)][ - 2[x/(x-1)^2)] - [x/(x-1)]
-x(x+1)/(x-1)^3

View other answers by:bugzpodder|Soarer|ali|kstahmer