Let's solve each step-by-step. --- ## 1. **Power Series for** \( f(x) = \frac{x(1+x)}{(1-x)^2} \) **centered at 0, and interval of convergence** Rewrite \( f(x) \): \[ f(x) = \frac{x(1+x)}{(1-x)^2} = \frac{x}{(1-x)^2} + \frac{x^2}{(1-x)^2} \] Recall: \[ \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1}, \quad |x| < 1 \] So, \[ \frac{x}{(1-x)^2} = x \cdot \sum_{n=1}^{\infty} n x^{n-1} = \sum_{n=1}^{\infty} n x^{n} \] \[ \frac{x^2}{(1-x)^2} = x^2 \cdot \sum_{n=1}^{\infty} n x^{n-1} = \sum_{n=1}^{\infty} n x^{n+1} = \sum_{k=2}^{\infty} (k-1)x^k \] (Replace \(n+1\) by \(k\), so \(n = k-1\), \(k\) starts at 2.) Now sum: \[ f(x) = \sum_{n=1}^{\infty} n x^n + \sum_{k=2}^{\infty} (k-1)x^k \] Combine into one sum: \[ = x + \sum_{n=2}^{\infty} n x^n + (n-1)x^n = x + \sum_{n=2}^{\infty} (n + n-1)x^n = x + \sum_{n=2}^{\infty} (2n-1) x^n \] **Final Power Series:** \[ \boxed{ f(x) = x + \sum_{n=2}^{\infty} (2n-1)x^n } \] **Interval of convergence:** \( |x| < 1 \) --- ## 2. **Maclaurin Series for** \( f(x) = \int_0^{7x}(e^{-t^2} - 1)dt \) Expand \(e^{-t^2}\) in Maclaurin series: \[ e^{-t^2} = \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!} \] So, \[ e^{-t^2} - 1 = \sum_{n=1}^{\infty} \frac{(-1)^n t^{2n}}{n!} \] Integrate term-by-term from \(0\) to \(7x\): \[ f(x) = \int_0^{7x} \sum_{n=1}^{\infty} \frac{(-1)^n t^{2n}}{n!} dt = \sum_{n=1}^{\infty} \frac{(-1)^n}{n!} \int_0^{7x} t^{2n} dt = \sum_{n=1}^{\infty} \frac{(-1)^n}{n!} \left[ \frac{t^{2n+1}}{2n+1} \right]_0^{7x} \] \[ = \sum_{n=1}^{\infty} \frac{(-1)^n}{n!} \frac{(7x)^{2n+1}}{2n+1} \] **Final Maclaurin Series:** \[ \boxed{ f(x) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n!} \frac{(7x)^{2n+1}}{2n+1} } \] --- ## 3. **Radius of Convergence for** \( \sum_{n=0}^\infty \frac{(n!)k^n x^n}{(kn)!} \) Let’s use the **ratio test**: \[ a_n = \frac{n! k^n x^n}{(kn)!} \] \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)! k^{n+1} x^{n+1}}{(k(n+1))!} \cdot \frac{(kn)!}{n! k^n x^n} \right| \] \[ = \lim_{n \to \infty} \left| \frac{(n+1) k x (kn)!}{(k(n+1))(k(n+1)-1)\cdots(kn+1) (n)!} \right| \] But as \(n\) grows, \((kn)!\) grows much faster than \(n!\), so the series converges **for all real \(x\)**. **Radius of convergence:** \(\boxed{\infty}\) --- ## 4. **Interval of Convergence for** \[ \sum_{n=1}^\infty \frac{(-1)^{n+1}3 \cdot 7 \cdot 11 \cdots (4n-1)(x-3)^n}{4^n} \] Write the general term: \[ a_n = \frac{(-1)^{n+1} \prod_{k=1}^n (4k-1) (x-3)^n}{4^n} \] Use the root test or ratio test. The dominant growth is in \((x-3)^n/4^n\), since the product factor is like \(n!\) (but less than exponential). Use ratio test: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{n+2} \prod_{k=1}^{n+1} (4k-1)(x-3)^{n+1}/4^{n+1}}{(-1)^{n+1} \prod_{k=1}^n (4k-1)(x-3)^n/4^n} \right| \] \[ = \lim_{n \to \infty} \left| \frac{(4(n+1)-1)(x-3)}{4} \right| = \lim_{n \to \infty} \left| \frac{(4n+3)(x-3)}{4} \right| \to \infty \text{ for any } |x-3| > 0 \] So the series **converges only at \(x = 3\)**. **Interval of convergence:** \(\boxed{x = 3}\) --- ### **Summary of Answers** 1. \(\frac{x(1+x)}{(1-x)^2} = x + \sum_{n=2}^{\infty} (2n-1)x^n\), \(|x| < 1\) 2. \(f(x) = \sum_{n=1}^\infty \frac{(-1)^n}{n!} \frac{(7x)^{2n+1}}{2n+1}\) 3. Radius of convergence: \(\infty\) 4. Interval of convergence: \(x = 3\) only Let me know if you want more details on any part!

Certainly! Let's go through each problem step-by-step with detailed calculations, explanations, and the final answers. --- ## 1. Power Series for \( f(x) = \frac{x(1 + x)}{(1 - x)^2} \) ### Step 1: Rewrite \( f(x) \) \[ f(x) = \frac{x(1 + x)}{(1 - x)^2} = \frac{x}{(1 - x)^2} + \frac{x^2}{(1 - x)^2} \] ### Step 2: Recall the known series expansion \[ \frac{1}{(1 - x)^2} = \sum_{n=1}^\infty n x^{n-1} \quad \text{for } |x| < 1 \] ### Step 3: Expand each term - For \( \frac{x}{(1 - x)^2} \): \[ \frac{x}{(1 - x)^2} = x \cdot \sum_{n=1}^\infty n x^{n-1} = \sum_{n=1}^\infty n x^{n} \] - For \( \frac{x^2}{(1 - x)^2} \): \[ \frac{x^2}{(1 - x)^2} = x^2 \cdot \sum_{n=1}^\infty n x^{n-1} = \sum_{n=1}^\infty n x^{n+1} \] Change index to align powers: \[ \sum_{n=1}^\infty n x^{n+1} = \sum_{k=2}^\infty (k-1) x^{k} \] (Since \(k = n + 1\), when \(n = 1\), \(k=2\).) ### Step 4: Write \(f(x)\) as a sum \[ f(x) = \sum_{n=1}^\infty n x^{n} + \sum_{k=2}^\infty (k-1) x^{k} \] Express both sums over \(n\): \[ f(x) = n=1 \text{ term} + \sum_{n=2}^\infty n x^{n} + \sum_{n=2}^\infty (n-1) x^{n} \] Note that the \(n=1\) term from the first sum is \(1 \cdot x^1 = x\). Now combine the sums for \(n \geq 2\): \[ f(x) = x + \sum_{n=2}^\infty [n + (n - 1)] x^n = x + \sum_{n=2}^\infty (2n - 1) x^n \] ### **Final Power Series:** \[ \boxed{ f(x) = x + \sum_{n=2}^\infty (2n - 1) x^n } \] ### **Interval of convergence:** Since the series originated from the expansion of \(1/(1-x)^2\), which converges for \(|x| < 1\), the power series for \(f(x)\) also converges for: \[ \boxed{ |x| < 1 } \] --- ## 2. Maclaurin Series for \(f(x) = \int_0^{7x} (e^{-t^2} - 1) dt\) ### Step 1: Expand \( e^{-t^2} \) \[ e^{-t^2} = \sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{n!} \] ### Step 2: Express \( e^{-t^2} - 1 \) \[ e^{-t^2} - 1 = \sum_{n=1}^\infty \frac{(-1)^n t^{2n}}{n!} \] ### Step 3: Integrate term-by-term from \(0\) to \(7x\) \[ f(x) = \int_0^{7x} \left( \sum_{n=1}^\infty \frac{(-1)^n t^{2n}}{n!} \right) dt \] Interchanging sum and integral (valid within convergence): \[ f(x) = \sum_{n=1}^\infty \frac{(-1)^n}{n!} \int_0^{7x} t^{2n} dt \] Calculate the integral: \[ \int_0^{7x} t^{2n} dt = \left[ \frac{t^{2n + 1}}{2n + 1} \right]_0^{7x} = \frac{(7x)^{2n + 1}}{2n + 1} \] ### **Final series:** \[ \boxed{ f(x) = \sum_{n=1}^\infty \frac{(-1)^n}{n!} \frac{(7x)^{2n + 1}}{2n + 1} } \] ### **Interval of convergence:** The series for \(e^{-t^2}\) converges for all real \(t\), so the integral converges for all \(x\). The power series in \(x\) converges for all \(x\) because the sum involves powers of \(x\) multiplied by a factorial term, which ensures convergence for all real \(x\). **Therefore:** \[ \boxed{ \text{Radius of convergence} = \infty } \] --- ## 3. Radius of convergence of \( \sum_{n=0}^\infty \frac{n! k^n x^n}{(kn)!} \) ### Step 1: Write the general term \[ a_n = \frac{n! \, k^n x^n}{(kn)!} \] ### Step 2: Use the ratio test \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)! \, k^{n+1} x^{n+1}}{(k(n+1))!} \times \frac{(kn)!}{n! \, k^n x^n} \right| \] Simplify: \[ = \lim_{n \to \infty} \left| \frac{(n+1) \, k \, x \times (kn)!}{(k(n+1))!} \right| \] Now, note: \[ (k(n+1))! = (k n + k)! = \prod_{j=1}^{k} (k n + j) \] Thus, the ratio becomes: \[ \left| (n+1) \times k \times x \times \frac{1}{\prod_{j=1}^k (k n + j)} \right| \] As \(n \to \infty\), \(\prod_{j=1}^k (k n + j) \sim (k n)^k\), which grows faster than \(n!\). Therefore, the limit goes to zero for any finite \(x\). ### **Conclusion:** The series converges for **all \(x\)**, so: \[ \boxed{ \text{Radius of convergence} = \infty } \] --- ## 4. Interval of convergence for \[ \sum_{n=1}^\infty \frac{(-1)^{n+1} \, 3 \times 7 \times 11 \times \cdots \times (4n - 1) \times (x - 3)^n}{4^n} \] ### Step 1: Analyze the coefficients The numerator: \[ \prod_{k=1}^n (4k - 1) \] This is a product of linear terms, which grows roughly like \( (4n)! \) but more precisely, it's a polynomial sequence. ### Step 2: Use the ratio test to find the convergence at endpoints The ratio test: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] where \[ a_n = \frac{\prod_{k=1}^n (4k - 1) \times (x - 3)^n}{4^n} \] Calculate: \[ \frac{a_{n+1}}{a_n} = \frac{\prod_{k=1}^{n+1} (4k - 1) \times |x - 3|^{n+1} / 4^{n+1}}{\prod_{k=1}^n (4k - 1) \times |x - 3|^n / 4^n} = \frac{(4(n+1) - 1) \times |x - 3|}{4} = \frac{(4n + 3) |x - 3|}{4} \] As \(n \to \infty\): \[ L = \lim_{n \to \infty} \frac{(4n + 3)|x - 3|}{4} = \infty \text{ unless } |x - 3| = 0 \] Thus, the series **converges only when \(x=3\)**. --- # **Final Summaries:** | Part | Power Series / Expression | Interval / Radius of Convergence | |---|---|---| | 1 | \( f(x) = x + \sum_{n=2}^\infty (2n - 1) x^n \) | \( |x| < 1 \) | | 2 | \( f(x) = \sum_{n=1}^\infty \frac{(-1)^n}{n!} \frac{(7x)^{2n+1}}{2n+1} \) | \( \text{All real } x \) (radius \( \infty \)) | | 3 | \( \text{Radius of convergence} = \infty \) | All \(x\) | | 4 | Series converges only at \(x=3\) | \(x=3\) only | --- Let me know if you'd like further clarifications or details!