Proving that ${n+3\choose 3} =\frac{n+2}{2}\sum_{k=1}^{n+1}\csc^2\frac{k\pi}{n+2}$

Using parts of this question and answers $$\sum_{k=1}^{n+1}\frac{1}{\sin^2\left(\frac{k\pi}{n+2}\right)}= \sum_{k=1}^{n+1}\left(1+\cot^2\left(\frac{k\pi}{n+2}\right)\right)=\\ n+1+\sum_{k=1}^{n+1}\left(\cot^2\left(\frac{k\pi}{n+2}\right)\right)=n+1+\frac{(n+1)n}{3}=\frac{(n+3)(n+1)}{3}$$

Note: Jack's answer, with a link to wikipedia, touches this question/answers as well.

Proof using linear algebra :

Initial expression

$${m+3\choose 3} =\frac{m+2}{2}\sum_{k=1}^{m+1}\sin\left(\tfrac{k\pi}{m+2}\right)^{-2},\tag{1}$$

can be transformed, by expressing ${m+3\choose 3}=\dfrac16(m+3)(m+2)(m+1)$ into :

$$\dfrac13(m+3)(m+1) =\sum_{k=1}^{m+1}\sin\left(\tfrac{k\pi}{m+2}\right)^{-2},\tag{1'}$$

Let us separate cases $m$ even and $m$ odd.

First case : $m:=2n$.

In this case, expression (1') is equivalent (due to the property $\sin(\pi-x)=\sin(x)$) to :

$$\dfrac13(2n+3)(2n+1) =1+2\sum_{k=1}^{n}\sin\left(\tfrac{k\pi}{2(n+1)}\right)^{-2},\tag{1''}$$

which is itself equivalent to :

$$\sum_{k=1}^n \dfrac{1}{4 \sin\left(\tfrac{k\pi}{2(n+1)}\right)^2}= \dfrac{n(n+2)}{6}\tag{2}$$

Consider the $n \times n$ tridiagonal matrix

$$D_n=\begin{pmatrix}2&-1&0&0&0&0&\cdots\\ -1&2&-1&0&0&0&\cdots\\ 0&-1&2&-1&0&0&\cdots\\ 0&0&-1&2&-1&0&\cdots\\ 0&0&0&-1&2&-1&\cdots\\ 0&0&0&0&-1&2&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots& \end{pmatrix}\tag{3}$$

The eigenvalues of $D_n$ are (see Eigenvalues of tridiagonal symmetric matrix with diagonal entries 2 and subdiagonal entries 1) :

$$\lambda_k:=4 \sin\left(\tfrac{k\pi}{2(n+1)}\right)^2, \ \ k=1,2,\cdots n\tag{4}$$

Remark : matrix $D_n$ is the classical discretized version of the second derivative (with a minus sign) : see the "Neumann case" in https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative.

Let us now consider $E_n:=D_n^{-1}.$

For example :

$$E_6:=D_6^{-1}=\dfrac17\begin{pmatrix} 6&5&4&3&2&1\\ 5&10&8&6&4&2\\ 4&8&12&9&6&3\\ 3&6&9&12&8&4\\ 2&4&6&8&10&5\\ 1&2&3&4&5&6 \end{pmatrix}\tag{5}$$

In these matrices, we just need the general fact that the diagonal entries of $D_n$ obey the following formula :

$$E_{n,kk}=\tfrac{1}{n+1}k(n+1-k)\tag{6}.$$

(see Remark 1 below for a proof of (6))

Therefore :

$$trace(E)=\tfrac{1}{n+1}(\sum_{k=1}^n k(n+1-k)) \tag{7}$$

$$=\tfrac{1}{n+1}((n+1)\tfrac{n(n+1)}{2}-\tfrac{n(n+1)(2n+1)}{6})=\tfrac{n(n+2)}{6} \tag{8} $$

i.e., the RHS of (2).

Besides,

$$trace(E)=\sum \dfrac{1}{\lambda_k}$$ gives the LHS of (2), establishing this relationship.

Remarks :

1) Proof of formula (6). We need a first result which is rather easy to establish by recurrence : $\det(D_n)=n+1$. The cofactor associated with $E_{kk}$ is obtained by cancelling the $k$th row and $k$th column of $A_n$, giving a diagonal-by-block determinant with blocks $A_{k-1}$ and $A_{n-k)}$, whose determinants are resp. $k$ and $n-k+1$. This minor is to be divided by the determinant of $A_n,$ i.e., $n+1.$

2) In (8), we have used the classical formulas for the sum of the $n$ first integers, resp. the squares of the $n$ first integers.

3) A slightly different version of matrix $D$ and its remarkable inverse can be found as second example in the following question : Looking for examples of Discrete / Continuous complementary approaches This is a typical case of an ill-conditionned matrix where a small perturbation (on the bottom right coefficient) gives a very different inverse.

4) Matrix $D_n$ is connected to Chebyshev polynomials of the 2nd kind $U_n$ ; see the answer https://math.stackexchange.com/q/1770607

Second case : $m=2n-1$

Relationship (1') (to be established) is :

$$\dfrac16 n(n+1) =\sum_{k=1}^{n}\left(4\sin\left(\tfrac{k\pi}{2n+1}\right)\right)^{-2},\tag{9}$$

Let us consider now the $n \times n$ tridiagonal matrix :

$$\Delta_n:=\begin{pmatrix}2&1&&&&\\1&2&1&&&\\&1&2&1&&\\&&\ddots&\ddots&\ddots\\&&&1&2&1\\&&&&1&3\end{pmatrix}$$

(please note the bottom right exceptional entry $3$).

It is not difficult to prove that $\Delta_n$ has the following eigenvalues :

$$\mu_k=4 \sin^2\left(\frac{k\pi}{2n+1}\right),\qquad k=1\dots n$$

One can prove as well that $\det \Delta_n=2n+1$.

The inverse $\Gamma_n:=\Delta_n^{-1}$ is a structured matrix whose diagonal elements are :

$$\Gamma_{n,kk}=k\dfrac{1}{2n+1}k(2n+1-2k)=k-2\dfrac{1}{2n+1}k^2\tag{10}$$

(proof using the same arguments as those used in Remark 1 above.)

Using (10), the trace of $\Gamma$ is

$$\sum_{k=1}^n k-2\dfrac{1}{2n+1}\sum_{k=1}^n k^2=\dfrac{n(n+1)}{6}$$

which is the LHS of (9), completing the proof.