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

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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
$$
R = e^{x^4} - 1.
$$
$$ R = e^{x^4} - 1 $$

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