summation - Sum of Arithmetic-Geometric Series

So I am having trouble getting the sum of the series:

$1 + 2\left(\frac32\right) + 3\left(\frac{3^2}{2^2}\right) + ... + k\left(\frac{3^{k-1}}{2^{k-1}}\right)$

I cant figure out for the life of me how to figure out the sum of this.

Answer

Let
$$S = 1+a+a^2+\cdots + a^k = {1-a^{k+1}\over 1-a}$$
Differentiating for $a$ we get
$$S^\prime = 1 + 2a + 3a^2 + \cdots + ka^{k-1} = {a^k(ka-k-1)+1\over (a-1)^2}$$
It is then obvious that for $a={3\over 2}$ we get the wanted sum, that is
$$1+2\bigl({3\over 2}\bigr) + \cdots + k\bigl({3\over 2}\bigr)^{k-1}=\bigl({3\over 2}\bigr)^k(2k-4)+4$$