1. **Problem:** Find the coefficients $a_0$, $a_1$, and $a_2$ in the power series expansion $$\frac{1}{(1 - x)^2} = \sum_{n=0}^\infty a_n x^n$$ for $|x| < 1$. **Step 1:** Recall that the binomial series expansion for $(1 - x)^{-k}$ is $$\sum_{n=0}^\infty {n + k - 1 \choose n} x^n.$$ For $k=2$, the coefficient is: $$a_n = {n + 2 - 1 \choose n} = {n + 1 \choose n} = n + 1.$$ **Step 2:** Calculate the first three coefficients: $$a_0 = 0 + 1 = 1,$$ $$a_1 = 1 + 1 = 2,$$ $$a_2 = 2 + 1 = 3.$$ **Answer:** The coefficients are $1, 2,$ and $3$. 2. **Problem:** Let $f : [-3, \infty) \to [-8, \infty)$ be defined by $f(x) = x^2 + 6x + 1$. Determine which statement about $f$ is true. **Step 1:** Complete the square for $f(x)$: $$f(x) = x^2 + 6x + 1 = (x+3)^2 - 9 + 1 = (x+3)^2 - 8.$$ **Step 2:** The domain is $[-3, \infty)$, so $x+3 \geq 0$. The function $f$ is then increasing on this domain, so it is one-to-one. **Step 3:** The range is $[-8, \infty)$, so $f$ maps onto $[-8, \infty)$, meaning it is onto that interval. **Step 4:** Find the inverse $f^{-1}(y)$. Set $y = (x+3)^2 - 8$, so $$(x+3)^2 = y + 8,$$ $$x + 3 = \sqrt{y + 8},$$ since $x \geq -3$, $$x = -3 + \sqrt{y + 8}.$$ **Answer:** The true statement is (E): $f$ is one-to-one and onto, with inverse $$f^{-1}(x) = -3 + \sqrt{x + 8}.$$ 3. **Problem:** Let $p, q$ be primes with $2 < p < q$. For $M$ the set of positive integers $n$ such that $n^5$ is divisible by $p^3$ and by $64q^{11}$, find the smallest $n$ in $M$. **Step 1:** Write prime factorization: - Divisible by $p^3$ means $p^3 | n^5$ - Divisible by $64q^{11}$ means $2^6 q^{11} | n^5$ (since $64 = 2^6$). **Step 2:** For primes dividing $n$, their exponents in $n^5$ must be at least as large as required. For prime $p$, Let exponent in $n$ be $a_p$, then $$5a_p \geq 3 \Rightarrow a_p \geq \frac{3}{5}.$$ The smallest integer exponent $a_p$ is 1. **Step 3:** For prime 2, exponent $a_2$: $$5 a_2 \geq 6 \Rightarrow a_2 \geq \frac{6}{5} = 1.2,$$ so smallest integer exponent is 2. **Step 4:** For prime $q$, exponent $a_q$: $$5 a_q \geq 11 \Rightarrow a_q \geq \frac{11}{5} = 2.2,$$ so smallest integer exponent is 3. **Step 5:** So minimal $n$ is $$n = 2^{2} p^{1} q^{3} = 4 p q^{3}.$$ **Answer:** The least integer in $M$ is $4 p q^{3}$. **Final answers:** 10. Coefficients are $(1, 2, 3)$ — option (B). 11. True statement is (E). 12. Least integer in $M$ is $4 p q^{3}$ — option (C).