如何计算三角求和式$S_n = \sum_{k=1}^n\frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2n}\right)}$？

Let's work through this problem step by step, using the provided hints and polynomial root/coefficient relationships (Vieta's formulas) to find a closed-form solution.

Step 1: Use the Trigonometric Identity from the Hint #

First, recall the core identity that connects cosecant and cotangent:
$$\frac{1}{\sin^2 x} = 1 + \cot^2 x$$
Applying this to every term in our sum, we get exactly the form noted in your first hint:
$$S_n = \sum_{k=1}^n \left(1 + \cot^2\left(\frac{(2k+1)\pi}{2n}\right)\right) = n + \sum_{k=1}^n \cot^2\left(\frac{(2k+1)\pi}{2n}\right)$$
Now our problem reduces to calculating the sum of cotangent squares, then adding $n$ to get $S_n$.

Step 2: Link the Angles to a Polynomial Equation #

Consider the angles $\theta_k = \frac{(2k+1)\pi}{2n}$ for $k=1,2,...,n$. Notice that multiplying by $2n$ gives:
$$2n\theta_k = (2k+1)\pi$$
Taking the cosine of both sides, we use the fact that $\cos((2k+1)\pi) = -1$, so:
$$\cos(2n\theta_k) + 1 = 0$$

To turn this into a polynomial, let $x = \cot\theta$. We know $\cos\theta = \frac{x}{\sqrt{1+x^2}}$ and $\sin\theta = \frac{1}{\sqrt{1+x^2}}$, so the complex exponential form of $\cos\theta + i\sin\theta$ is $\frac{x+i}{\sqrt{1+x^2}}$. Raising this to the $2n$-th power gives:
$$\left(\frac{x+i}{\sqrt{1+x^2}}\right){2n} = e^{i2n\theta} = -1$$
Multiply both sides by $(1+x²⁾n$ to eliminate the denominator, resulting in the polynomial equation:
$$(x+i)^{2n} + (1+x²⁾n = 0$$

Step 3: Expand and Simplify the Polynomial #

Expand $(x+i)^{2n}$ using the binomial theorem:
$$(x+i)^{2n} = \sum_{m=0}^{2n} \binom{2n}{m}x^{2n-m}im$$
Split this into real and imaginary parts. Since the right-hand side $-(1+x²⁾n$ is real, the imaginary part must equal 0, and the real part must match $-(1+x²⁾n$.

Focus on the real part and substitute $y = x^2 = \cot^2\theta$ to get an $n$-th degree polynomial in $y$:
$$\sum_{k=0}^n \binom{2n}{2k}(-1)^{n-k}yk + \sum_{k=0}^n \binom{n}{k}y^k = 0$$

Step 4: Use Vieta's Formulas to Find the Sum of Roots #

The roots of this polynomial are exactly $y_k = \cot^2\left(\frac{(2k+1)\pi}{2n}\right)$ for $k=1,...,n$. Using Vieta's formula (sum of roots = $-\frac{\text{coefficient of } y^{n-1}}{\text{coefficient of } y^n}$):

• The leading coefficient (for $y^n$) is $\binom{2n}{2n}(-1)^0 + \binom{n}{n} = 1 + 1 = 2$.
• The coefficient for $y^{n-1}$ is $\binom{2n}{2(n-1)}(-1)^1 + \binom{n}{n-1} = -\binom{2n}{2} + n = -n(2n-1) + n = -2n(n-1)$.

Calculating the sum of the roots:
$$\sum_{k=1}^n y_k = -\frac{-2n(n-1)}{2} = n(n-1)$$

Step 5: Compute the Final Sum $S_n$ #

Substitute back into our initial identity:
$$S_n = n + \sum_{k=1}^n \cot^2\left(\frac{(2k+1)\pi}{2n}\right) = n + n(n-1) = n^2$$

Note: The second hint's guess of $S_n = \frac{n^2}{4}$ is incorrect—our derivation shows the correct result is $S_n = n^2$, which checks out with small values of $n$ (e.g., $n=1$ gives $1$, $n=2$ gives $4$, $n=3$ gives $9$).

内容的提问来源于stack exchange，提问作者Eric