I can't understand a how the podium is built in my problem So we are going to build a podium with blocks with the measurements $1 \times 1 × 1$ inch the question is as follows. 
The top layer should measure $2 \times 2$ inches and then each layer should project $1/2$ inch on all four sides relative to the layer above (so that the podium will be staircase-shaped)
and the sum for n layer is $\dfrac13n^3+\dfrac32n^2+\dfrac{13}6n$
before I used the equation I thought it was 4 blocks in the top layer and 12 blocks the next layer, but when I tried to confirm that with the equation, it showed for layer 1 4 blocks and for layer 2 13 blocks. So did I misunderstand how to build the podium or did i miss calculate the equation? 

Since the podium projects $\frac{1}{2}$ inch on each of the four sides, going from the top layer of $2 \times 2$ inches to the next layer would increase the size by $\frac{1}{2} + \frac{1}{2} = 1$ inch in its width, and the same in its height, to give that each dimension is now $2 + 1 = 3$ inches, i.e., the second layer dimensions are $3$ inches by $3$ inches. This results in $3 \times 3 = 9$ blocks for the second layer, for a total of $4 + 9 = 13$ blocks for the $2$ layers combined, just as the equation for the sum gives.
Note also when you wrote "for layer 2 13 blocks" that the equation is for the sum of the blocks up to layer $n$, not just the number of blocks in the $n$'th layer.
To confirm, have $f(n)$ be the total # of blocks up to layer $n$ to get
$$\begin{equation}\begin{aligned}
f(n) & = \frac{1}{3}n^3 + \frac{3}{2}n^2 + \frac{13}{6}n \\
& = \frac{2n^3 + 9n^2 + 13n}{6}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, $f(1) = \frac{2 + 9 + 13}{6} = \frac{24}{6} = 4$, and $f(2) = \frac{16 + 36 + 26}{6} = \frac{78}{6} = 13$.
