How can I find the total number of figures in a puzzle where they seem to overlap? The problem is as follows:

In the following sequence of figures. Which is the total number of
  right trapezoids which do make up $\textrm{figure 10}$?

The figure is below:

The alternatives given are:


*

*66

*60

*56

*64

*72


What I tried to do is to account for the number of trapezoids:
$\textrm{In figure 1:}$


*

*There is just $1$


$\textrm{In figure 2:}$
This is some tricky:
I thought that there are $4$


*

*$2$ make up a small individual unit

*$1$ accounts for the other two

*$1$ accounts for the all of them
$\textrm{In figure 3:}$


*

*$3$ are consisting of only one figure.

*$2$ are two figures.

*$1$ has three figures.

*$1$ accounts for the three trapezoids in horizontal arrangement and the first oblique bar next to them.

*$1$ accounts for the previous set plus the second oblique bar next to them.
Therefore in the figure are $8$
$\textrm{In figure 4:}$


*

*$4$ are sets of 1 figure.

*$3$ are sets of 2 figures.

*$2$ are sets of 3 figures.

*$1$ is a set of 4 figures.

*$1$ is a set of the previous 4 figures in horizontal direction plus the first oblique bar appearing next to the trapezoids.

*$1$ is the previous set and the second oblique bar.

*$1$ is the previous set and the third oblique bar.
This accounts for $13$ figures.
So in total the series is comprised of.
$\textrm{1, 4, 8, 13,...}$
From this series I found that the recursion formula is stated as:
$$T_{n}=\frac{1}{2}n^{2}+\frac{3}{2}n-1$$
By replacing with the fourth term it seems to check.
$$T_{4}=\frac{1}{2}\left( 4 \right )^{2}+\frac{3}{2}\left( 4 \right )-1=8+6-1= 13$$
So if what it is being asked is figure number $10$ then by replacing in the previous equation:
$$T_{10}=\frac{1}{2}\left( 10 \right )^{2}+\frac{3}{2}\left( 10 \right )-1=50+15-1= 64$$
Therefore I would choose $64$ as the number of figures in the $\textrm{10th figure}$. I'm not very certain if is it correct?
The answer appears within the alternatives but the process to find the figures was very tedious and it took me a long time to find it, needless to say that to find the recursive formula was another problem as well but eased due the fact which I had some idea of how to find it, as a result.
Is there any other method to speed up the calculations or to find an answer in this situation or to avoid miscounting or double counting the figures?
 A: This looks correct to me.  As for the more general question, I would approach the problem by asking 'how can I make up a right trapezoid from these figures?' This breaks naturally into two pieces: trapezoids that use the 'slanting' shapes on the right of the figure, and ones that don't.
If I'm going to use any of the shapes on the right then it's clear I have to use all the ones on the left, and I have to use the ones on the right from left to right; thus, there are as many ways of doing this as there are trapezoids on the right of the figure, i.e. $n-1$.
On the other hand, if I'm using on the sub-trapezoids on the left, then it has to be a contiguous block of them (e.g., the second through the fourth would be valid, but the first and third would not), but any such contiguous block will do; you should be able to see that the number of contiguous blocks is the same as the number of ways of choosing a number $i$ (the 'bottom block') between 1 and $n$ and a number $j$ (the 'top block') between $1$ and $i$; there are exactly $n(n+1)/2$ of these.
So summing up, I get a total of $n(n+1)/2+n-1 = \frac12n^2+\frac32n-1$ possible trapezoids, matching your answer.
A: I think you got it.
Counting the stack of $n$ trapezoids in the left part of each diagram is the $n$th triangular number -- normally denoted as $T_n$ but I'll denote it $t_n$ so as not to clash with your notation. The tenth triangular number $t_{10} = 55.$
As additional explanation, let's look at the fourth diagram in your question. Let's label the stack of short, fat trapezoids (looks like a geometrical stack of pancakes) on the left part of that diagram $A,B,C,D$ from top to bottom.
You have $4$ trapezoids with the topmost trapezoid at the top ($A$ by itself, $A+B$, $A+B+C$, and $A+B+C+D$). You'd have $3$ trapezoids with the second-highest trapezoid on top ($B$ by itself, $B+C$, and $B+C+D$). And so on. So the total number of trapezoids, of any height in this stack, is $4+3+2+1 = 10$, which is the fourth triangular number.
Similarly, in the tenth diagram, it would be $10 + 9 + 8 + ... + 2 + 1 = t_{10} = 55$.
A different but completely equivalent way to look at it is that there is $1$ trapezoid that is stacked $10$ high, $2$ trapezoids that are stacked $9$ high, $3$ that are stacked $8$ high ... and $10$ that are stacked $1$ high. It still adds up to $55$.
Since the side pieces extend only the trapezoid that represents the entire stack, you add $n-1$ of those. (From the diagrams you have, the third diagram has two additional trapezoids from the side pieces, hence the $n-1$ formula.)
Adding these two disjoint sets of trapezoids gives your answer $55+(10-1)=64$.
A: ETA: I've added a somewhat shoddy diagram to illustrate what I mean.  I've also changed the notation somewhat to be (a bit) more logical.
I'm not sure this is much simpler than what you did, but I numbered the vertices down the left-hand side $B_{0, 0}, B_{1, 0}, \ldots, B_{n, 0}$, then rightward from $B_{0, 0}$ were $B_{0, 1}, B_{0, 2}, \ldots, B_{0, n}$, and rightward from $B_{n, 0}$ were $B_{n, 1}, B_{n, 2}, \ldots, B_{n, n}$.  Finally, down the leftmost diagonal, between $B_{0, 1}$ and $B_{n, 1}$, were $B_{1, 1}, B_{2, 1}, \ldots, B_{n-1, 1}$.

Hopefully, it'll be clear that the right trapezoid must contain two vertices of the form $B_{i, 0}$ and $B_{j, 0}$ (without loss of generality, assume $i < j$), since the right angles only live there.  The other two vertices must then be $B_{i, k}$ and $B_{j, k}$, for some value of $k$.
Then, there are only two basic classes of right trapezoids:


*

*If $k = 1$, then $i$ and $j$ can be any values such that $0 \leq i < j \leq n$.

*If $2 \leq k \leq n$, then $i$ and $j$ must be $0$ and $n$, respectively.
There are $\binom{n+1}{2}$ of the former, and $n-1$ of the latter, which gives us a count of
$$
\binom{n+1}{2}+n-1 = \frac{n^2+3n-2}{2}
$$
which gives us $64$ when $n = 10$, as expected.
