We seek to evaluate
$$\sum_{l=0}^m (-4)^l {m\choose l} {2l\choose l}^{-1}
\sum_{k=0}^n \frac{(-4)^k}{2k+1} {n\choose k} {2k\choose k}^{-1}
{k+l\choose l}.$$
We start with the inner term and use the Beta function identity
$$\frac{1}{2k+1} {2k\choose k}^{-1}
= \int_0^1 x^k (1-x)^k \; dx.$$
We obtain
$$\int_0^1 [z^l]
\sum_{k=0}^n {n\choose k} (-4)^k x^k (1-x)^k \frac{1}{(1-z)^{k+1}}
\; dx
\\ = [z^l] \frac{1}{1-z}
\int_0^1 \left(1-\frac{4x(1-x)}{1-z}\right)^n \; dx
\\ = [z^l] \frac{1}{(1-z)^{n+1}}
\int_0^1 ((1-2x)^2-z)^n \; dx
\\ = \sum_{q=0}^l {l-q+n\choose n}
[z^q] \int_0^1 ((1-2x)^2-z)^n \; dx
\\ = \sum_{q=0}^l {l-q+n\choose n}
{n\choose q} (-1)^q \int_0^1 (1-2x)^{2n-2q} \; dx
\\ = \sum_{q=0}^l {l-q+n\choose n}
{n\choose q} (-1)^q
\left[-\frac{1}{2(2n-2q+1)} (1-2x)^{2n-2q+1}\right]_0^1
\\ = \sum_{q=0}^l {l-q+n\choose n}
{n\choose q} (-1)^q \frac{1}{2n-2q+1}.$$
Now we have
$$ {l-q+n\choose n} {n\choose q} (-1)^q \frac{1}{2n-2q+1}
\\ = \mathrm{Res}_{z=q}
\frac{(-1)^n}{2n+1-2z}
\prod_{p=0}^{n-1} (l+n-p-z) \prod_{p=0}^n \frac{1}{z-p}.$$
Residues sum to zero and since $\lim_{R\to\infty} 2\pi R \times R^n /
R / R^{n+1} = 0$ we may evaluate the sum using the negative of the
residue at $z=(2n+1)/2.$ We get
$$\frac{1}{2} (-1)^n
\prod_{p=0}^{n-1} (l+n-p-(2n+1)/2)
\prod_{p=0}^n \frac{1}{(2n+1)/2-p}
\\ = (-1)^n
\prod_{p=0}^{n-1} (2l+2n-2p-(2n+1))
\prod_{p=0}^n \frac{1}{2n+1-2p}
\\ = (-1)^n
\prod_{p=0}^{n-1} (2l-2p-1)
\frac{2^n n!}{(2n+1)!}
\\ = (-1)^n \frac{1}{2l+1}
\prod_{p=-1}^{n-1} (2l-2p-1)
\frac{2^n n!}{(2n+1)!}
\\ = (-1)^n \frac{2^n n!}{(2n+1)!} \frac{1}{2l+1}
\prod_{p=0}^{n} (2l-2p+1)
\\ = (-1)^n \frac{2^{2n+1} n!}{(2n+1)!} \frac{1}{2l+1}
\prod_{p=0}^{n} (l+1/2-p)
\\ = (-1)^n \frac{2^{2n+1} n! (n+1)!}{(2n+1)!} \frac{1}{2l+1}
{l+1/2\choose n+1}.$$
We obtain for our sum
$$(-1)^n 2^{2n+1} {2n+1\choose n}^{-1}
\sum_{l=0}^m (-4)^l {m\choose l} \frac{1}{2l+1} {2l\choose l}^{-1}
{l+1/2\choose n+1}.$$
We now work with the remaining sum without the factor in front. We
obtain
$$\int_0^1 [z^{n+1}] \sqrt{1+z}
\sum_{l=0}^m {m\choose l} (-4)^l x^l (1-x)^l (1+z)^l \; dx
\\ = [z^{n+1}] \sqrt{1+z}
\int_0^1 (1-4x(1-x)(1+z))^m \; dx
\\ = [z^{n+1}] \sqrt{1+z}
\int_0^1 \sum_{q=0}^m {m\choose q} (1-2x)^{2m-2q}
(-1)^q (4x(1-x))^q z^q \; dx
\\ = \sum_{q=0}^m {m\choose q} {1/2\choose n+1-q}
\int_0^1
(1-2x)^{2m-2q}
(-1)^q (4x(1-x))^q \; dx
\\ = \sum_{q=0}^m {m\choose q} {1/2\choose n+1-q}
\int_0^1
(1-2x)^{2m}
\left(1-\frac{1}{(1-2x)^2}\right)^q \; dx
\\ = \sum_{q=0}^m {m\choose q} {1/2\choose n+1-q}
\sum_{p=0}^q {q\choose p} (-1)^p \int_0^1 (1-2x)^{2m-2p} \; dx
\\ = \sum_{q=0}^m {m\choose q} {1/2\choose n+1-q}
\sum_{p=0}^q {q\choose p} (-1)^p \frac{1}{2m-2p+1}.$$
Re-writing then yields
$$\sum_{p=0}^m (-1)^p \frac{1}{2m-2p+1}
\sum_{q=p}^m {m\choose q} {1/2\choose n+1-q} {q\choose p}.$$
Observe that
$${m\choose q} {q\choose p} =
\frac{m!}{(m-q)! \times p! \times (q-p)!}
= {m\choose p} {m-p\choose m-q}$$
so that we find
$$\sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1}
\sum_{q=p}^m {m-p\choose m-q} {1/2\choose n+1-q}
\\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1}
\sum_{q=0}^{m-p} {m-p\choose m-p-q} {1/2\choose n+1-p-q}
\\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1}
\sum_{q=0}^{m-p} {m-p\choose q} {1/2\choose n+1-p-q}.$$
Continuing we obtain
$$\sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1}
\sum_{q=0}^{m-p} {m-p\choose q} [z^{n+1-p}] z^q \sqrt{1+z}
\\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1}
[z^{n+1-p}] \sqrt{1+z} \sum_{q=0}^{m-p} {m-p\choose q} z^q
\\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1}
[z^{n+1-p}] (1+z)^{m-p+1/2}
\\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1}
{m-p+1/2\choose n+1-p}
\\ = (-1)^m \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2p+1}
{p+1/2\choose n+1-m+p}
\\ = (-1)^m
\sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2} \frac{1}{m-n-1/2}
{p-1/2\choose n+1-m+p}
\\ = (-1)^m \frac{1}{2m-2n-1}
\sum_{p=0}^m {m\choose p} (-1)^p
{p-1/2\choose n+1-m+p}.$$
Concluding with a closed form we establish at last
$$(-1)^m \frac{1}{2m-2n-1}
\sum_{p=0}^m {m\choose p} (-1)^p [z^{n+1-m}] z^{-p} (1+z)^{p-1/2}
\\ = (-1)^m \frac{1}{2m-2n-1} [z^{n+1-m}] (1+z)^{-1/2}
\sum_{p=0}^m {m\choose p} (-1)^p z^{-p} (1+z)^p
\\ = (-1)^m \frac{1}{2m-2n-1}
[z^{n+1-m}] (1+z)^{-1/2} \left(1-\frac{1+z}{z}\right)^m
\\ = \frac{1}{2m-2n-1} [z^{n+1}] (1+z)^{-1/2}.$$
We finish by re-introducing the factor in front to obtain
$$(-1)^n 2^{2n+1} {2n+1\choose n}^{-1} \frac{1}{2m-2n-1}
{-1/2\choose n+1}
\\ = (-1)^n 2^{2n+1} {2n+1\choose n}^{-1} \frac{1}{2m-2n-1}
\frac{1}{(n+1)!} \prod_{q=0}^{n} (-1/2 -q)
\\ = (-1)^n 2^{n} {2n+1\choose n}^{-1} \frac{1}{2m-2n-1}
\frac{1}{(n+1)!} \prod_{q=0}^{n} (-1 -2q)
\\ = 2^{n} {2n+1\choose n}^{-1} \frac{1}{2n+1-2m}
\frac{1}{(n+1)!} \prod_{q=0}^{n} (1 +2q)
\\ = 2^{n} {2n+1\choose n}^{-1} \frac{1}{2n+1-2m}
\frac{1}{(n+1)!} \frac{(2n+1)!}{2^n n!}.$$
Yes indeed this is
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2n+1-2m}.}$$
Here I have chosen to document the simple steps as well as the
complicated ones to aid all types of readers.
Here is a tentative proof:
We first notice that by Kummer theorem $\binom{2n}{2m} \equiv \binom{n}{m} \bmod 2$, so that it suffices to show that $$ S_q(n) \equiv T_q(n) \pmod 2$$ where
\begin{align*}
S_q(n):=&\sum_k \binom{k}{n-k}\binom{n+3q}{2k+2q+1}\\
T_q(n):=&\binom{n+3q}{2q-1}.
\end{align*}
From here, the symbol $\sim$ means has the same parity as.
We have
$\binom{a+1}{b+1}\sim \binom{a}{b+1}+\binom{a}{b}$ but also $\binom{a+2}{b+2}\sim \binom{a+2}{b}+\binom{a}{b}$.
We have
\begin{align*} T_q(n+2)&\sim \binom{n+3q+2}{2q-1} \sim \binom{n+3q}{2q-1}+\binom{n+3q}{2q-3}\\
&\sim T_q(n) +\binom{n+3+3(q-1)}{2(q-1)-1} \\
&\sim T_q(n) +T_{q-1}(n+3) \\
T_q(n+3)&\sim T_{q+1}(n+2)+ T_{q+1}(n)
\end{align*}
On the other hand, we have
\begin{align*}S_q(n+3)& \sim \sum_k \binom{k}{n+3-k}\binom{n+3q+3}{2k+2q+1}\\
&\sim \sum_k \binom{k}{n+2-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&\sim \sum_k \binom{k-1}{n+2-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&+ \sum_k \binom{k-1}{n+1-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&\sim \sum_k \binom{k-1}{n+2-(k-1)}\binom{n+2+3q+1}{2(k-1)+2q+3}\\
&+ \sum_k \binom{k-1}{n+1-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&\sim \sum_k \binom{k-1}{n+2-(k-1)}\binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)+1}\\
&+ \sum_k \binom{k-1}{n+1-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
& \sim \sum_k \binom{k-1}{n+2-(k-1)}\binom{n+2+3(q+1)}{2(k-1)+2(q+1)+1}\\
&+\sum_k\binom{k-1}{n+2-(k-1)}\binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)-1}\\
&+\sum_k \binom{k-1}{n+1-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&\sim S_{q+1}(n+2) +\sum_k\binom{k}{n+2-k}\binom{n+3(q+1)}{2k+2(q+1)-1}\\
&+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2) \\&+\sum_k\binom{k}{n+2-k}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\\
&+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+\sum_k\binom{k-1}{n+2-k}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\\
&+\sum_k\binom{k-1}{n-(k-1)}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\\
&+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+ S_{q+1}(n) \\&+\sum_k\binom{k-1}{n+2-k}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+ S_{q+1}(n)+2\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+ S_{q+1}(n) \\&+\sum_k\binom{k-1}{n+2-k}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+ S_{q+1}(n)+2\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
S_q(n+3)&\sim S_{q+1}(n+2)+ S_{q+1}(n)
\end{align*}
So the parity of both $ S_q(n)$ and $ T_q(n)$ satisfy the same third order linear recurrence on $n$. Then in order to show that $ S_q(n) \sim T_q(n)$, it suffices to show that $ S_q(0) \sim T_q(0)$, $ S_q(1) \sim T_q(1)$ and $ S_q(2) \sim T_q(2)$.
$S_q(0)-T_q(0)$ is clearly an even integer since
$$S_q(0)-T_q(0)= \binom{3q}{q+1}-\binom{3q}{q-1}=2 \binom{3q+1}{q-1}.$$
We have $T_q(1)-S_q(1)= \binom{3q+1}{q+2}-\binom{3q+1}{q-2}$. After some calculation we find $$T_q(1)-S_q(1)= 2\frac{5q+7}{3q+3}\binom{3q+3}{q-1}$$ which is an even integer since \begin{align*}\binom{3q+3}{q-1}(5q+7) &\equiv \binom{3q+3}{q-1}(2q+4)\pmod {3q+3}\\
&\equiv-\binom{3q+3}{q-1}(q-1)\pmod {3q+3}\\
&\equiv-\binom{3q+2}{q-2}(3q+3)\pmod {3q+3}\\
&\equiv 0\pmod {3q+3}\end{align*}
Also, after some calculation, it can be shown that
\begin{align*}
T_q(2)-S_q(2) &= \binom{3q+2}{q+3}-\binom{3q+2}{q-1}-\binom{3q+2}{q-3}\\
&= 2 \binom{3q+5}{q-1} \frac{54+301q+258q^2+59q^3}{(3q+5)(3q+4)(3q+3)}.
\end{align*}
Then, to complete the proof, we need to show that
$$ \binom{3q+5}{q-1} (54+301q+258q^2+59q^3)\equiv 0 \pmod {(3q+5)(3q+4)(3q+3)}.$$
We have $$ (54+301q+258q^2+59q^3) \equiv (q-1)(66+47q+5q^2)\pmod {(3q+5)(3q+4)(3q+3)}.$$
Then $$ \binom{3q+5}{q-1} (54+301q+258q^2+59q^3) \equiv (3q+5)\binom{3q+4}{q-2}(66+47q+5q^2)\pmod {(3q+5)(3q+4)(3q+3)}.$$
Then it suffices to show that
$$ \binom{3q+4}{q-2}(66+47q+5q^2)\equiv 0 \pmod {(3q+4)(3q+3)}.$$
But $3q+4$ and $3q+3$ are coprime, so we need to show that
$$ \binom{3q+4}{q-2}(66+47q+5q^2)\equiv 0 \pmod {3q+4}$$
and
$$ \binom{3q+4}{q-2}(66+47q+5q^2)\equiv 0 \pmod {3q+3}.$$
That is
$$ \binom{3q+4}{q-2}(14-q^2)\equiv 0 \pmod {3q+4} \tag1$$
and
$$ \binom{3q+4}{q-2}(24-q-q^2)\equiv 0 \pmod {3q+3}.\tag2$$
(1) is equivalent to
\begin{align*} \frac{3q+4}{q-2}\binom{3q+3}{q-3}(14-q^2)&\equiv 0 \pmod {3q+4} \\
\binom{3q+3}{q-3}(14-q^2)&\equiv 0 \pmod {q-2}\\
10\binom{3q+3}{q-3}&\equiv 0 \pmod {q-2}\\
10\frac{q-2}{3q+4}\binom{3q+4}{q-2}&\equiv 0 \pmod {q-2}\\
10\binom{3q+4}{q-2}&\equiv 0 \pmod {3q+4}.\\
\end{align*}
But the last line holds true indeed, because we know from here that
\begin{align*} \binom{3q+4}{q-2}&\equiv 0 \pmod {\frac{3q+4}{\gcd(3q+4,q-2)}}\\
&\equiv 0 \pmod {\frac{3q+4}{\gcd(10,q-2)}}\end{align*}
There remains to show the validity of (2). We have
$$ \binom{3q+4}{q-2}(24-q-q^2)= (3q+3)\binom{3q+4}{q-4} \frac{3q+4}{(q-2)(q-3)}(24-q-q^2)$$
so that we need to show that
$$ \binom{3q+2}{q-4}(3q+4)(24-q-q^2)\equiv 0\pmod {(q-2)(q-3)}. $$
That is
$$ \binom{3q+2}{q-4}(3q+4)(24-q-q^2)\equiv 0\pmod {q-2} $$
and
$$ \binom{3q+2}{q-4}(3q+4)(24-q-q^2)\equiv 0\pmod {q-3}. $$
That is
$$ 180\binom{3q+2}{q-4}\equiv 0\pmod {q-2} $$
and
$$ 156 \binom{3q+2}{q-4}\equiv 0\pmod {q-3}. $$
That is
$$ 180\frac{(q-2)(q-3)}{(3q+4)(q-2)}\binom{3q+4}{q-2}\equiv 0\pmod {q-2} $$
and
$$ 156 \frac{q-3}{3q+3}\binom{3q+3}{q-3}\equiv 0\pmod {q-3}. $$
That is
$$ 180(q-3)\binom{3q+4}{q-2}\equiv 0\pmod {(3q+4)(3q+3)} $$
and
$$ 156\binom{3q+3}{q-3}\equiv 0\pmod {3q+3}. $$
That is
$$ 180(q-3)\binom{3q+4}{q-2}\equiv 0\pmod {3q+4} \tag3$$
and
$$ 180(q-3)\binom{3q+4}{q-2}\equiv 0\pmod {3q+3} \tag4$$
and
$$ 156\binom{3q+3}{q-3}\equiv 0\pmod {3q+3}. \tag5$$
For (5) it holds true because we have $156=13\cdot 12$ a multiple of $12$ and from here, we have
\begin{align*}\binom{3q+3}{q-3}&\equiv 0\pmod {\frac{3q+3}{\gcd(3q+3,q-3)}} \\
&\equiv 0\pmod {\frac{3q+3}{\gcd(12,q-3)}} \end{align*}
For (3) it holds true because we have $180=18\cdot 10$ a multiple of $10$ and from here, we have
\begin{align*}\binom{3q+4}{q-2}&\equiv 0\pmod {\frac{3q+4}{\gcd(3q+4,q-2)}} \\
&\equiv 0\pmod {\frac{3q+2}{\gcd(10,q-2)}} \end{align*}
We have seen that there exist an integer $K$, such that $\binom{3q+3}{q-3}=K\frac{3q+3}{\gcd(12,q-3)}$. Now, with the same argument from here, we have
\begin{align*}\binom{3q+3}{q-2}&\equiv 0\pmod {\frac{3q+3}{\gcd(3q+3,q-2)}} \\
&\equiv 0\pmod {\frac{3q+3}{\gcd(9,q-2)}} \end{align*}
so there exists an integer $L$ such that $\binom{3q+3}{q-2}=L\frac{3q+3}{\gcd(9,q-2)}$.
Then
\begin{align*} 180(q-3)\binom{3q+4}{q-2}&=180(q-3)\left(\binom{3q+3}{q-2}+\binom{3q+3}{q-3}\right)\\
&= (3q+3)\left( \frac{180L(q-3)}{\gcd(9,q-2)}+\frac{180K(q-3)}{\gcd(12,q-3)}\right). \end{align*}
The second factor on the right hand side is clearly an integer and this establishes the validity of (4). The proof is finished here.
Best Answer
The result follows from the equality of two different expressions for the Chebyshev polynomials of the first kind. We have $$ \begin{aligned} T_N(x)&=\sum_{j\ge0}\binom{N}{2j}(x^2-1)^j x^{N-2j}\\ &=\frac{1}{2}\sum_{r\ge0}(-1)^r\frac{N}{N-r}\binom{N-r}{r}(2x)^{N-2r}, \end{aligned} $$ where the first equality holds for $N\ge0$ and the second for $N\ge1$. Expanding the binomial factor in the first expression gives $$ \begin{aligned} &\sum_{j\ge0}\binom{N}{2j}\sum_{r=0}^j\binom{j}{r}(-1)^r x^{2j-2r}x^{N-2j}\\ &\quad=\sum_{r\ge0}(-1)^rx^{N-2r}\sum_{j\ge r}\binom{N}{2j}\binom{j}{k}\\ &\quad=\sum_{r\ge0}(-1)^rx^{N-2r}\sum_{j\ge 0}\binom{N}{2r+2j}\binom{r+j}{r}. \end{aligned} $$ Comparing coefficients yields $$ \sum_{j\ge 0}\binom{N}{2r+2j}\binom{r+j}{r}=\frac{1}{2}\frac{N}{N-r}\binom{N-r}{r}2^{N-2r} $$ Setting $N=2n$ and $r=n-k$ gives your identity.
Of course for this to be a proof, we really have to prove that the two expressions for $T_N(x)$ hold. We define $T_N(x)$ by the condition $\cos(N\theta)=T_N(\cos\theta)$. The first expression for $T_N(x)$ follows by taking the real parts of $$ \cos(N\theta)+i\sin(N\theta)=e^{iN\theta}=\sum_{k=0}^N\binom{N}{k}(i\sin(\theta))^k\cos^{N-k}\theta $$ and recognizing that $(i\sin\theta)^{2j}=(\cos^2\theta-1)^j$.
The factor $\frac{N}{N-r}\binom{N-r}{r}$ in the second expression is the number of $r$-matchings on $C_N$, the cycle graph of $N$ vertices. Equivalently, it's the number of ways of placing $r$ non-overlapping dominoes on the edges of an $N$-gon. What does this have to do with expressing $\cos(N\theta)$ as a polynomial in $\cos\theta$? The idea is to add powers of $2\cos\theta=(e^{i\theta}+e^{-i\theta})$ up to the $N^\text{th}$ power, with coefficients chosen so the only the $e^{iN\theta}$ and $e^{-iN\theta}$ terms survive, and then multiply by $\frac{1}{2}$ to get $\cos(N\theta)$. To eliminate the unwanted terms, we use the principle of inclusion-exclusion, as follows. Represent a term in the expansion of $(e^{i\theta}+e^{-i\theta})^N$ by the sequence of signs in the exponent. So the term $e^{i\theta}e^{i\theta}e^{-i\theta}e^{i\theta}$ in the expansion of $(e^{i\theta}+e^{-i\theta})^4$ would be represented by the sign sequence $++-+$. We want to keep the terms $+++\ldots+$ and $---\ldots-$, and discard everything else. Define $S_j$ to the the set of sequences in which a plus at position $j$ is followed by a minus at position $j+1$, where $j$ ranges from $0$ to $N-1$ and $j+1$ is computed $\mod N$ (so that the sequence is considered to be wrapped on a circle). Since the terms $e^{i\theta}$ and $e^{-i\theta}$ at positions $j$ and $j+1$ cancel, the sum of the terms corresponding to sequences in $S_j$ is $(e^{i\theta}+e^{-i\theta})^{N-2}$. So, from $(e^{i\theta}+e^{-i\theta})^N$, we subtract, for each $j$, the quantity $(e^{i\theta}+e^{-i\theta})^{N-2}$. But if a term has a sequence in which $+$ is immediately followed by $-$ at two different positions, say $j$ and $k$, that term will have been subtracted twice, and therefore needs to be added back in. This requires adding $(e^{i\theta}+e^{-i\theta})^{N-4}$ for every such pair $j$, $k$. By the principle of inclusion--exclusion, we continue in this way, alternately adding and subtracting the terms $(e^{i\theta}+e^{-i\theta})^{N-2r}$ corresponding to sequences in $S_{j_1}\cap S_{j_2} \cap S_{j_3}\cap\ldots\cap S_{j_r}$. It only remains to determine how many non-empty intersections there are of $r$ sets. There is only one condition we need to worry about: if $+$ at $j$ is followed by $-$ at $j+1$, then it certainly is not the case that $+$ is followed by $-$ at positions $j+1$ and $j+2$, so any intersection containing $S_j\cap S_{j+1}$ is empty. This is precisely the non-overlapping domino condition, and the second expression for $T_N(x)$ follows.