Use the definition of a Taylor series to find the first four nonzero terms of the series for $f(x)$ centered at the given value of a. (Enter your answers as a comma-separated list.) \[ f(x)=\frac{7}{1+x}, \quad a=2 \]

Step 1 ：Given the function \(f(x)=\frac{7}{1+x}\), we are asked to find the first four nonzero terms of the Taylor series centered at \(a=2\).

Step 2 ：The Taylor series of a function about a point can be found using the formula: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]

Step 3 ：First, we need to find the derivatives of the function. The derivative of \(f(x)\) is \(f'(x) = -\frac{7}{(1+x)^2}\), the second derivative is \(f''(x) = \frac{14}{(1+x)^3}\), and the third derivative is \(f'''(x) = -\frac{42}{(1+x)^4}\).

Step 4 ：Next, we substitute \(a=2\) into these derivatives to find the values of \(f(a)\), \(f'(a)\), \(f''(a)\), and \(f'''(a)\). We find that \(f(a) = \frac{7}{3}\), \(f'(a) = -\frac{7}{9}\), \(f''(a) = \frac{14}{27}\), and \(f'''(a) = -\frac{42}{81}\).

Step 5 ：Finally, we substitute these values into the Taylor series formula to find the first four nonzero terms of the series. The first four nonzero terms of the Taylor series for the function \(f(x)=\frac{7}{1+x}\) centered at \(a=2\) are \(\frac{7}{3}\), \(\frac{14}{9} - \frac{7x}{9}\), \(\frac{7(x - 2)^2}{27}\), and \(-\frac{7(x - 2)^3}{81}\).

Step 6 ：\(\boxed{\text{Final Answer: } \frac{7}{3}, \frac{14}{9} - \frac{7x}{9}, \frac{7(x - 2)^2}{27}, -\frac{7(x - 2)^3}{81}}\)