\( R = \sum _ { n = 1 } ^ { \infty } \frac { x ^ { 4 n } } { n ! } \)

We start with the series \[ R = \sum_{n=1}^{\infty} \frac{x^{4n}}{n!}. \] **Step 1. Recognize a related Taylor series** Recall that the Taylor series expansion of the exponential function is \[ e^y = \sum_{n=0}^{\infty} \frac{y^n}{n!}. \] If we let \(y = x^4\), the series becomes \[ e^{x^4} = \sum_{n=0}^{\infty} \frac{x^{4n}}{n!}. \] **Step 2. Adjust the series to match \(R\)** Notice that the series for \(R\) starts at \(n=1\) instead of \(n=0\). The series for \(e^{x^4}\) includes the term when \(n=0\), which is \[ \frac{x^{4 \cdot 0}}{0!} = \frac{1}{1} = 1. \] Thus, we can write \[ e^{x^4} = 1 + \sum_{n=1}^{\infty} \frac{x^{4n}}{n!}. \] This shows that \[ \sum_{n=1}^{\infty} \frac{x^{4n}}{n!} = e^{x^4} - 1. \] **Step 3. Conclude the result** Therefore, the given series can be written in closed form as \[ R = e^{x^4} - 1. \]