Let's solve each part step by step: --- ### **Part 1:** **Use the binomial series to find the first four nonzero terms in the Taylor expansion for the given function at \( a \).** Given: \( f(x) = (1 + 2x)^5 \), \( a = 0 \) #### **Step 1: Binomial Series Expansion** The binomial theorem for any integer \( n \) is: \[ (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \] For \( f(x) = (1 + 2x)^5 \), substitute \( x \) with \( 2x \): \[ (1 + 2x)^5 = \sum_{k=0}^{5} \binom{5}{k} (2x)^k \] #### **Step 2: Write the first four nonzero terms** Let’s find the first four terms: - **For \( k = 0 \):** \(\binom{5}{0}(2x)^0 = 1\) - **For \( k = 1 \):** \(\binom{5}{1}(2x)^1 = 5 \cdot 2x = 10x\) - **For \( k = 2 \):** \(\binom{5}{2}(2x)^2 = 10 \cdot 4x^2 = 40x^2\) - **For \( k = 3 \):** \(\binom{5}{3}(2x)^3 = 10 \cdot 8x^3 = 80x^3\) #### **Step 3: Final Answer** The first four nonzero terms are: \[ \boxed{1 + 10x + 40x^2 + 80x^3} \] --- ### **Part 2:** **Use the binomial series to find the first four nonzero terms in the Taylor expansion for the given function at \( a \).** Given: \( f(x) = (1 + 2x)^{-5} \), \( a = 0 \) #### **Step 1: Binomial Series for Negative Exponent** \[ (1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k \] Where \(\binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{k!}\) For \( f(x) = (1 + 2x)^{-5} \): \[ (1 + 2x)^{-5} = \sum_{k=0}^{\infty} \binom{-5}{k} (2x)^k \] #### **Step 2: Compute the first four terms** - **For \( k=0 \):** \(\binom{-5}{0}(2x)^0 = 1\) - **For \( k=1 \):** \(\binom{-5}{1}(2x)^1 = (-5) \cdot 2x = -10x\) - **For \( k=2 \):** \(\binom{-5}{2}(2x)^2 = \frac{(-5)(-6)}{2!} \cdot 4x^2 = \frac{30}{2} \cdot 4x^2 = 15 \cdot 4x^2 = 60x^2\) - **For \( k=3 \):** \(\binom{-5}{3}(2x)^3 = \frac{(-5)(-6)(-7)}{3!} \cdot 8x^3\) Calculate numerator: \((-5)(-6)(-7) = -210\) Denominator: \(3! = 6\) So, \( \frac{-210}{6} \cdot 8x^3 = -35 \cdot 8x^3 = -280x^3 \) #### **Step 3: Final Answer** The first four nonzero terms are: \[ \boxed{1 - 10x + 60x^2 - 280x^3} \] --- ### **Part 3:** **Find the sum of the series:** \[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{3^n n} \] Recall the Taylor Series for \( \ln(1+x) \) at \( x=0 \): \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n \] Here, we have \( x = \frac{1}{3} \): \[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \left( \frac{1}{3} \right)^n = \ln\left(1 + \frac{1}{3}\right) = \ln\left(\frac{4}{3}\right) \] So, \[ \boxed{\ln\left(\frac{4}{3}\right)} \] --- ### **Summary of Final Answers:** 1. \( (1 + 2x)^5 \):                                                         \[ \boxed{1 + 10x + 40x^2 + 80x^3} \] 2. \( (1 + 2x)^{-5} \): \[ \boxed{1 - 10x + 60x^2 - 280x^3} \] 3. \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{3^n n} \): \[ \boxed{\ln\left(\frac{4}{3}\right)} \]