Problem 15. (7 points)
Find the first five non-zero terms of power serles representation centered at $x=0$ for the function below.
\[
f(x)=\frac{2 x^{3}}{(x-2)^{2}}
\]

Answer: $f(x)=\square+\square+\square+\square+\square+\cdots$
What is the radius of convergence?
Answer: $R=$

Problem 16. (7 points)

Step 1 ：Rewrite the function in a form that is easier to expand into a power series: \(f(x)=\frac{2 x^{3}}{(x-2)^{2}} = 2x \cdot \frac{x^{2}}{(x-2)^{2}}\)

Step 2 ：Use the formula for the geometric series to rewrite the fraction as a power series: \(\frac{1}{(1-x)^{n}} = \sum_{k=0}^{\infty} \binom{n+k-1}{k} x^{k}\)

Step 3 ：In our case, we have n=2 and x is replaced by x/2. So we get: \(\frac{x^{2}}{(x-2)^{2}} = \sum_{k=0}^{\infty} \binom{2+k-1}{k} \left(\frac{x}{2}\right)^{k}\)

Step 4 ：Multiply this series by 2x to get the power series representation of the function: \(f(x) = 2x \cdot \sum_{k=0}^{\infty} \binom{2+k-1}{k} \left(\frac{x}{2}\right)^{k}\)

Step 5 ：The first five non-zero terms of this series are: \(f(x) = 2x + 3x^{2} + 2x^{3} + \frac{2}{3}x^{4} + \frac{1}{6}x^{5} + \cdots\)

Step 6 ：The radius of convergence R is the value for which the series converges. In this case, the series converges for |x/2|<1, so \(\boxed{R=2}\)