1. **State the problem:** We want to find the coefficients $a_0$, $a_1$, and $a_2$ in the power series expansion of $$\left(\frac{1}{1-x}\right)^2 = \sum_{n=0}^\infty a_n x^n$$ where $|x| < 1$. 2. **Recall the known series:** The geometric series sum is $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$ for $|x| < 1$. 3. **Square the series:** Squaring both sides gives $$\left(\frac{1}{1-x}\right)^2 = \left(\sum_{n=0}^\infty x^n\right)^2 = \sum_{n=0}^\infty a_n x^n$$ where $a_n$ are the coefficients we want. 4. **Use Cauchy product formula:** The coefficients $a_n$ of the squared series are given by $$a_n = \sum_{k=0}^n 1 \cdot 1 = n+1$$ since each term is the sum of the products of coefficients of the original series which are all 1. 5. **Calculate specific coefficients:** - $a_0 = 0+1 = 1$ - $a_1 = 1+1 = 2$ - $a_2 = 2+1 = 3$ 6. **Summary:** The coefficients are $$a_0 = 1, \quad a_1 = 2, \quad a_2 = 3.$$