We need to evaluate the following limit:

$\lim_{n \to \infty} \sum_{i=1}^n \frac{7\pi^2 i}{4n^2} \cos\left(\frac{7\pi i}{2n}\right)$

This expression can be interpreted as a Riemann sum. To proceed, let's rewrite the sum in a form that makes it more evident:

1. Identify the form of the Riemann sum:

   The sum can be written as:

   $\sum_{i=1}^n f\left(\frac{i}{n}\right) \frac{1}{n}$

   where:

   $f(x) = \frac{7\pi^2 x}{4} \cos\left(\frac{7\pi x}{2}\right)$

   and $x = \frac{i}{n}$.

2. Transition to the integral:

   As $n \to \infty$, the sum becomes a definite integral over the interval $[0, 1]$:

   $\lim_{n \to \infty} \sum_{i=1}^n \frac{7\pi^2 i}{4n^2} \cos\left(\frac{7\pi i}{2n}\right) = \int_0^1 \frac{7\pi^2 x}{4} \cos\left(\frac{7\pi x}{2}\right) \, dx$

3. Evaluate the integral:

   The integral to evaluate is:

   $\int_0^1 \frac{7\pi^2 x}{4} \cos\left(\frac{7\pi x}{2}\right) \, dx$

   We can solve this by using integration by parts. Let:

   $u = x, \quad dv = \frac{7\pi^2}{4} \cos\left(\frac{7\pi x}{2}\right) dx$

   Then:

   $du = dx, \quad v = \frac{7\pi^2}{4} \cdot \frac{2}{7\pi} \sin\left(\frac{7\pi x}{2}\right) = \frac{7\pi}{14} \sin\left(\frac{7\pi x}{2}\right)$

   Using integration by parts:

   $\int u \, dv = uv - \int v \, du$

   Substituting the expressions:

   $\int_0^1 \frac{7\pi^2 x}{4} \cos\left(\frac{7\pi x}{2}\right) dx = \left[ \frac{7\pi}{14} x \sin\left(\frac{7\pi x}{2}\right) \right]_0^1 - \int_0^1 \frac{7\pi}{14} \sin\left(\frac{7\pi x}{2}\right) dx$

   Evaluating the first part at the bounds:

   $\frac{7\pi}{14} \cdot 1 \cdot \sin\left(\frac{7\pi}{2}\right) - \frac{7\pi}{14} \cdot 0 = 0$

   So the integral reduces to:

   $- \frac{7\pi}{14} \int_0^1 \sin\left(\frac{7\pi x}{2}\right) dx$

   Now, solve the remaining integral:

   $\int \sin\left(\frac{7\pi x}{2}\right) dx = -\frac{2}{7\pi} \cos\left(\frac{7\pi x}{2}\right)$

   Evaluating at the bounds $0$ and $1$:

   [