Summation with products of values with different powers

Question

For an exercise I am working on I have the following equation, $$ c\sum_{n=1}^{\infty}n\sum_{j=1}^{n}(\frac{\lambda}{2\mu})^{j}(\frac{\lambda}{3\mu})^{n-j}, $$ where $\lambda<\mu$. I have been trying to figure out how to take the first sum from $j=1$ to $n$, but I am unsure whether this is even possible. Is there a way that this could be done or is it likely I made a mistake in determining this equation.

Answer 1

Hint

Let $a=\frac{\lambda }{2 \mu }$ and $b=\frac{\lambda }{3 \mu }$ $$\sum_{j=1}^n a^j \,b^{n-j}={a^n}\sum_{j=1}^n \left(\frac{b}{a}\right)^{n-j}$$

$$k=\frac b a=\frac 23 \qquad \implies \quad\sum_{j=1}^n k^{n-j}=\frac {k ^{1-n }-k}{1-k }=2 \left(\left(\frac{3}{2}\right)^n-1\right)$$

The remaining is simple