Palindrome of fixed length
Imagine we build the DFA from top to bottom to recognize palindromes of length $2k$. As usual, we'll use the states to "remember" the input seen so far. The DFA will consist of three parts:


The states needed to represent the $2^k$ possible choices for the first $k$ characters. This will be a complete binary tree of depth $k$ with $2^k$ leaves and so will need $2^{k+1}-1$ nodes in total.
The states needed to process the remaining $k$ inputs. This will consist of $k-1$ layers, each of which contains $2^k$ nodes for a total of $(k-1)2^k$ nodes. For example, suppose we had $k=3$ and after part (1) we were in state $q_{011}$, i.e., having seen input $\mathtt{011}$. Then on input $\mathtt{0}$ we would transition to the fail state, since $\mathtt{0110}$ could not be part of a 6-character palindrome. If, on the other hand, we saw $\mathtt{1}$ we would change to a state on the next level, since the current input $\mathtt{0111}$ could be part of a palindrome.
Finally, we add an accept state and a reject state, with the appropriate transitions from the accept state to the reject state on either 0 or 1 and similarly from the reject state to itself on a 0 or a 1.


We will have $2^{k+1}-1$ states in stage (1), $(k-1)2^k$ states in stage (2) and the two states in stage (3) for a total of $(2^{k+1}-1)+(k-1)2^k+2=(k+1)2^k+1$ states.

It's not hard to use a Myhill-Nerode argument to show that the states are pairwise distinguishable and so this DFA is minimal.