1. **Problem statement:** Find the closed form of the power series \(\sum_{n=2}^\infty (n-1) x^{n+6}\) using the power series representation of \(\frac{1}{1-x}\). 2. **Recall the geometric series formula:** $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}, \quad |x|<1$$ and its derivative: $$\frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2} = \sum_{n=0}^\infty (n+1) x^n$$ 3. **Rewrite the given series:** $$\sum_{n=2}^\infty (n-1) x^{n+6} = x^6 \sum_{n=2}^\infty (n-1) x^n$$ 4. **Change the index to start from 0:** Let \(m = n-2\), then \(n = m+2\), so $$\sum_{n=2}^\infty (n-1) x^n = \sum_{m=0}^\infty (m+1) x^{m+2} = x^2 \sum_{m=0}^\infty (m+1) x^m$$ 5. **Use the derivative of geometric series:** $$\sum_{m=0}^\infty (m+1) x^m = \frac{1}{(1-x)^2}$$ 6. **Combine all parts:** $$\sum_{n=2}^\infty (n-1) x^{n+6} = x^6 \cdot x^2 \cdot \frac{1}{(1-x)^2} = \frac{x^8}{(1-x)^2}$$ --- 7. **Problem statement:** Find the Taylor series of \(f(x) = \frac{1}{x}\) centered at \(a = -1\). 8. **Rewrite function for expansion:** Express \(f(x) = \frac{1}{x} = \frac{1}{-1 + (x+1)} = \frac{1}{-1 + h}\) where \(h = x+1\). 9. **Rewrite as:** $$f(x) = \frac{1}{-1 + h} = -\frac{1}{1 - h}$$ 10. **Use geometric series expansion:** $$\frac{1}{1 - h} = \sum_{n=0}^\infty h^n = \sum_{n=0}^\infty (x+1)^n, \quad |x+1| < 1$$ 11. **Therefore:** $$f(x) = - \sum_{n=0}^\infty (x+1)^n = -1 - (x+1) - (x+1)^2 - (x+1)^3 - \dots$$ 12. **Domain of convergence:** The series converges for \(|x+1| < 1\), i.e., \(-2 < x < 0\). 13. **Given the problem states:** $$0 < x < \pi \quad \text{and} \quad f(x) = \pi \quad \text{for} \quad \pi < x < 2\pi$$ This suggests a piecewise definition outside the radius of convergence. --- **Final answers:** - For question 4: $$f(x) = \sum_{n=2}^\infty (n-1) x^{n+6} = \frac{x^8}{(1-x)^2}$$ - For question 5: $$\frac{1}{x} = - \sum_{n=0}^\infty (x+1)^n = -1 - (x+1) - (x+1)^2 - \dots, \quad |x+1| < 1$$