Let $n, k \in \mathbb N $ and $(X_1, \dotsc, X_{2n}) \sim \operatorname{Multinom}(2k, p_1, p_1, \dotsc, p_n, p_n)$.
Is there a closed form of
$$
P(X_1 = X_2, \dotsc, X_{2n-1} = X_{2n}) = \sum_{k_1 + \dotsc + k_n = k} P(X_1 = k_1, X_2 = k_1, \dotsc, X_{2n-1} = k_n, X_{2n} = k_n)
$$
EDIT:To clarify, I am NOT interested in $P(X_1 = X_2 = \dotsb = X_{2n})$. I just want each $X_{2i-1} = X_{2i}$. I know how to write it in a sum, i am looking for close form.
To simplify notation I shall use $q_1,...,q_{2n}$ instead of $p_1,p_1,...,p_n,p_n$.
By the definition of the multinomial distribution, \begin{align} &\mathbb P(X_1=k_1,...,X_{2n}=k_{2n})\\[3mm] \tag{1} &=\left\{\begin{array}~\displaystyle\frac{(2n)!}{k_1!...k_{2n}!}\,q_1^{k_1}\,...\,q_{2n}^{k_{2n}}\,,&\text{ when }k_1+...+k_{2n}=2n\,,\\0\,,&\text{ otherwise. }\end{array}\right. \end{align} The probability you seem to be interested in is then \begin{align} &\mathbb P(X_1=X_2,X_3=X_4,...,X_{2n-1}=X_{2n})\\ &=\sum_{k_1,...,k_n=1}^{2n}\mathbb P(X_1=k_1,X_2=k_1,X_3=k_2,X_4=k_2,...,X_{2n-1}=k_n,X_{2n}=k_n)\,. \end{align} In that sum only the terms with $2k_1+...+2k_n=2n$, or equivalently $k_1+...+k_n=n,$ have non zero probability (by (1)). Therefore, this can be written as \begin{align} &\sum_{k_1+...+k_n=n}\mathbb P(X_1=k_1,X_2=k_1,X_3=k_2,X_4=k_2,...,X_{2n-1}=k_n,X_{2n}=k_n)\\ &=\sum_{k_1+...+k_n=n}\frac{(2n)!}{k_1!k_1!...k_n!k_n!}p_1^{k_1}p_1^{k_1}...p_n^{k_n}p_n^{k_n}\\ &=\sum_{k_1+...+k_n=n}\frac{(2n)!}{n!n!}\Bigg(\frac{n!}{k_1!...k_n!}p_1^{k_1}...p_n^{k_n}\Bigg)^2\\ &={2n\choose n}\sum_{k_1+...+k_n=n}\Bigg(\frac{n!}{k_1!...k_n!}p_1^{k_1}...p_n^{k_n}\Bigg)^2\,. \end{align}