What kind of a sum is this?

Here's an approach using some calculus .....

Rewrite the expression as $\dfrac{2}{5} \sum_{n=1}^\infty n(\dfrac{3}{5})^{n-1}$ .

Now in general, for $|x| \lt 1$ , we have that

$\sum_{n=1}^\infty x^{n} = \dfrac{x}{x - 1}$ .

Taking the derivative of both sides, (the left side term-by-term), we have that

$\sum_{n=1}^\infty nx^{n-1} = \dfrac{1}{(1 - x)^{2}}$ .

So with $x = \dfrac{3}{5}$ our original expression becomes

$\dfrac{2}{5} * \dfrac{1}{(1 - \frac{3}{5})^{2}} = \dfrac{5}{2}$ .

Thus $a = 5, b = 2$ and $a + b = \boxed{7}$ .

Say there is some event that repeats with probability $\dfrac{3}{5}$ . That is, when it happens, there is a $\dfrac{3}{5}$ probability of it happening again.

Then the expected number of times this thing will happen given that it happens at least once is $1 + \dfrac{3}{5} + \dfrac{9}{25} \cdots = \dfrac{1}{1 - \dfrac{3}{5}} = \dfrac{1}{\dfrac{2}{5}} = \dfrac{5}{2}$ .

But the expected number of times can also be written another way. The probability that is happens once is $\dfrac{2}{5}$ , the probability that it happens twice is $\dfrac{3}{5} \cdot \dfrac{2}{5}$ , and in general the probability that it happens $n$ times is $(\dfrac{3}{5})^{n-1} \cdot \dfrac{2}{5} = \dfrac{3^{n-1} \cdot 2}{5^n}$ .

Now, we multiply the probability that this event happens $n$ times by $n$ , so each probability offers a certain "weight" to the sum. We get $n \cdot \dfrac{3^{n-1} \cdot 2}{5^n}$ . Thus, this value is another way to express the expected number of times this event will happen, and we get that the sum is $\dfrac{5}{2}$ , and $5+2=\boxed{7}$ .