No, because 2x/1 = 2x and 3x^2/(2x) = (3/2)x. The terms are not related by a constant factor.

Your written answer is incorrect. (And you really need to use parenthesis!)

Sn(x) = 1 + 2x + 3x^2 + ... + nx^{n - 1}

Now

1*Sn(x) = 1 + 2x + 3x^2 + ... + nx^{n - 1}

x*Sn(x) = x + 2x^2 + 3x^3 + ... + nx^n

So

(1 - x)*Sn(x) = 1 + x + x^2 + ... + x^{n - 1} - nx^n

Now 1 + x + x^2 + ... + x^{n - 1} is a geometric series with n terms, so the sum of this is:

(1 - x^n)/(1 - x)

Thus:

(1 - x)*Sn(x) = (1 - x^n)/(1 - x) - nx^n

Sn(x) = (1 - x^n)/(1- x)^2 - nx^n/(1 - x)

-Dan