Suppose that $\frac{2 x}{(4+x)}=\sum_{n=0}^{\infty} c_{n} x^{n}$.
Find the first few coefficients.
\[
\begin{array}{l}
c_{0}=\square \\
c_{1}=\square \\
c_{2}=\square \\
c_{3}=\square \\
c_{4}=\square
\end{array}
\]

Find the radius of convergence $R$ of the power series.
\[
R=\square
\]

Step 1 ：Given the equation \(\frac{2 x}{(4+x)}=\sum_{n=0}^{\infty} c_{n} x^{n}\), we are asked to find the first few coefficients and the radius of convergence of the power series.

Step 2 ：We can rewrite the given equation as a geometric series in the form of \(\frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio. In this case, \(a=2\) and \(r=-\frac{x}{4+x}\).

Step 3 ：We can find the coefficients \(c_n\) by expanding the geometric series and comparing the coefficients of the same powers of \(x\) on both sides of the equation. The expanded series is \(2 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{8} + \frac{x^4}{16} + O(x^5)\).

Step 4 ：By comparing the coefficients, we find that \(c_0=2\), \(c_1=-\frac{1}{2}\), \(c_2=\frac{1}{4}\), \(c_3=-\frac{1}{8}\), and \(c_4=\frac{1}{16}\).

Step 5 ：The radius of convergence \(R\) of a geometric series is given by \(|r|<1\). In this case, \(|-\frac{x}{4+x}|<1\). Solving this inequality, we find that \(R\) is \(x > -2\).

Step 6 ：So, the first few coefficients are \(\boxed{c_{0}=2}\), \(\boxed{c_{1}=-\frac{1}{2}}\), \(\boxed{c_{2}=\frac{1}{4}}\), \(\boxed{c_{3}=-\frac{1}{8}}\), and \(\boxed{c_{4}=\frac{1}{16}}\). The radius of convergence \(R\) is \(\boxed{x > -2}\).