How to find the $E(N)$ using $E(M)$ where the $M$ and $N$ follow slightly different scenarios 
An author sends his first manuscript to a large number of publishers, $C, D, E, ...$ , in turn, only
  approaching each one, after the first, if the one before has refused it. There is a constant probability
  $\frac{1}{4}$ that his manuscript will be accepted by each publisher approached. The random variable M is the 
  number of publishers approached, up to and including the one who accepts the manuscript. Write down the values of $E(M)$ and $Var(M)$
For his second manuscript the author decides to approach two other publishers, A then B, for each
  of whom the probability of acceptance is $\frac{1}{2}$, before he approaches $C, D, E, ...$ . The probability of 
  acceptance by each of these publishers remains $\frac{1}{4}$. The random variable N is the number of $A, B, C,
D, E, ...$ approached, up to and including the one who accepts this manuscript. Write down the first
  few terms of the series for $E(M)$ and for $E(N)$. By comparing corresponding terms, after the second,
  and using your value for $E(M)$, find $E(N)$.


What  I did : 
I got the first part correct.
For the first part, I realized that this is a Geometric distribution. So for a Geometric Distribution: 
$$E(M)=\frac{1}{p}=4$$
$$Var(M)=\frac{1-p}{p^2}=12$$

The problem is with this (second Manuscript), I wrote down the first few terms:
For M:
$$\frac{1}{4} , \frac {3}{4}\frac{1}{4} , \frac{3}{4}\frac{3}{4}\frac{1}{4},\frac{3}{4}\frac{3}{4}\frac{3}{4}\frac{1}{4},\frac{3}{4}\frac{3}{4}\frac{3}{4}\frac{3}{4}\frac{1}{4},...$$
For N:
$$\frac{1}{2} , \frac {1}{2}\frac{1}{2} , \frac{1}{2}\frac{1}{2}\frac{1}{4},\frac{1}{2}\frac{1}{2}\frac{3}{4}\frac{1}{4},\frac{1}{2}\frac{1}{2}\frac{3}{4}\frac{3}{4}\frac{1}{4},...$$

How to find $E(N)$ using this? I have no clue. Please help. They say by comparing, how can one find it by comparing?
 A: We do it in a somewhat different way that the one suggested.  If A accepts (probability $1/2$, then $N=1$. If A rejects but B accepts (probability $1/4$) then $N=2$. And if A and B both reject, then $N=M+2$, for we are back in the first scenario, except that we have gotten $2$ rejections already. The conditional expectation of $N$, given that there have been two rejections, is $E(M+2)$. It follows that
$$E(N)=\frac{1}{2}\cdot 1+\frac{1}{4}\cdot 2+\frac{1}{4}\cdot E(M+2).$$
We have $E(M+2)=E(M)+2$, and now we can use the answer for the first scenario to complete the calculation.
If you want to do it using series, write down the series for the expectation of $N$. You will see that the terms for $E(N)$ from $3$ on are closely related to the terms for $E(M)$. Multiply the $(k+2)$-th term by $4$, and subtract the $k$-th term of the series for $E(M)$. You will get a nice series.
Remark: We have done a conditional expectation calculation, in this case conditioning on the first two trial results. This kind of strategy is often useful.
