Find the sum $2\cdot\frac23+4\cdot\frac23\cdot\frac13+6\cdot\frac23\cdot\frac1{3^2}+8\cdot\frac23\cdot\frac1{3^3}+10\cdot\frac23\cdot\frac1{3^4}+...$

Question

My first thoughts was that this had something to do with infinite geometric series, but upon closer inspection I saw that this isn't a geometric series. I then tried to factor it into $$2\left(\frac23+\frac29+\frac2{27}+\dots\right)+2\left(\frac29+\frac2{27}+\frac2{81}+\dots\right)+2\left(\frac2{27}+\frac2{81}+\frac2{243}+\dots\right)+\dots$$But since this is infinite too, I am not sure how to proceed.

*this is not the original question, I was just able to deduce it to this. Let me know if there's an easier was to solve this:

You have been placed in the center of a maze. Each room is a shape and paths are lines. Every move, you randomly pick a path, leading you to a different room, independent of any previous moves. Once you reach a star room, you win. What is the expected number of moves you will make before winning?

Answer 1

Symmetry helps to solve the underlying puzzle (I think this is easier and less error prone than using the series).

There are only three states: $S_0$ means you are in the center, $S_1$ means you are in a square, $S_2$ means you are done (you have reached the star).

Let $E=E_0$ be the expected number of moves it takes given that you are in $S_0$ (so this is the answer you want).

Let $E_1$ be the expected number of moves it takes given that you are in a square.

Clearly $E_0=E_1+1$.

We also have $$E_1=\frac 23\times 1+\frac 13\times (E_0+1)=1+\frac 13\times E_0\implies E_0=3$$

and we are done.

As a minor variant, consider the first two moves. Either we are done (probability $\frac 23$) or we are back in the center (probability $\frac 13$). Thus $$E=\frac 23\times 2+\frac 13\times (E+2)\implies E=3$$

Note, the series you wrote in the title is $$\sum_{n=1}^{\infty}\frac {4n}{3^n}$$ which you can compute by differentiating $\frac 1{1-x}=\sum x^n$ and adjusting the terms a little (there are other ways to compute it, of course). I note that this also gives the value $3$ in the end.

Answer 2

To take you down the road you have already started, note that in the expression $$ 2\cdot\left(\frac23+\frac29+\frac2{27} + \cdots\right)+2\cdot\left(\frac29+\frac2{27}+\frac2{81} + \cdots\right)+2\cdot\left(\frac2{27}+\frac2{81} + \frac2{243} +\cdots\right)+\cdots $$ each of the brackets is a geometric series. And also, going from bracket to bracket, each is a third of the one before, so the whole sum is also a geometric series. Use what you know about geometric series to find the sum in the first bracket. Then use that each bracket is a third of the one before, and use what you know about geometric series to find the whole sum.