1. **Problem Statement:** Given the sequence $U_r = \frac{4(2r+7)}{(2r+1)(2r+3)(2r+5)}$ for $r \in \mathbb{Z}^+$, and a function $f(r) = \frac{A}{2r+1} + \frac{B}{2r+3}$ where $A,B$ are real numbers, find $A$ and $B$ such that $U_r = f(r) - f(r+1)$. 2. **Expressing $U_r$ as a telescoping difference:** We want $$U_r = f(r) - f(r+1) = \left(\frac{A}{2r+1} + \frac{B}{2r+3}\right) - \left(\frac{A}{2(r+1)+1} + \frac{B}{2(r+1)+3}\right).$$ Simplify the denominators: $$f(r+1) = \frac{A}{2r+3} + \frac{B}{2r+5}.$$ So, $$f(r) - f(r+1) = \frac{A}{2r+1} + \frac{B}{2r+3} - \frac{A}{2r+3} - \frac{B}{2r+5} = \frac{A}{2r+1} + \frac{B - A}{2r+3} - \frac{B}{2r+5}.$$ 3. **Equate to $U_r$ and find $A,B$:** We have $$U_r = \frac{4(2r+7)}{(2r+1)(2r+3)(2r+5)} = \frac{A}{2r+1} + \frac{B - A}{2r+3} - \frac{B}{2r+5}.$$ Multiply both sides by $(2r+1)(2r+3)(2r+5)$: $$4(2r+7) = A(2r+3)(2r+5) + (B - A)(2r+1)(2r+5) - B(2r+1)(2r+3).$$ 4. **Expand each term:** - $A(2r+3)(2r+5) = A(4r^2 + 16r + 15)$ - $(B - A)(2r+1)(2r+5) = (B - A)(4r^2 + 12r + 5)$ - $-B(2r+1)(2r+3) = -B(4r^2 + 8r + 3)$ Sum: $$4(2r+7) = A(4r^2 + 16r + 15) + (B - A)(4r^2 + 12r + 5) - B(4r^2 + 8r + 3).$$ 5. **Combine like terms:** Group coefficients of $r^2$, $r$, and constants: - Coefficient of $r^2$: $$4A + 4(B - A) - 4B = 4A + 4B - 4A - 4B = 0.$$ - Coefficient of $r$: $$16A + 12(B - A) - 8B = 16A + 12B - 12A - 8B = 4A + 4B.$$ - Constant term: $$15A + 5(B - A) - 3B = 15A + 5B - 5A - 3B = 10A + 2B.$$ 6. **Equate to left side:** Left side is $8r + 28$. So, $$0 \cdot r^2 + (4A + 4B)r + (10A + 2B) = 8r + 28.$$ Equate coefficients: $$4A + 4B = 8,$$ $$10A + 2B = 28.$$ 7. **Solve the system:** From first, $$4A + 4B = 8 \implies A + B = 2.$$ From second, $$10A + 2B = 28.$$ Express $B = 2 - A$ from first and substitute: $$10A + 2(2 - A) = 28 \implies 10A + 4 - 2A = 28 \implies 8A = 24 \implies A = 3.$$ Then, $$B = 2 - 3 = -1.$$ 8. **Check $U_r = f(r) - f(r+1)$ with $A=3$, $B=-1$:** $$f(r) = \frac{3}{2r+1} - \frac{1}{2r+3}.$$ 9. **Sum of $U_r$ from $r=1$ to $n$:** Given $$\sum_{r=1}^n U_r = \frac{4}{5} - \frac{3}{2n+3} + \frac{1}{2n+5}.$$ 10. **Check convergence of the infinite sum:** Take limit as $n \to \infty$: $$\lim_{n \to \infty} \sum_{r=1}^n U_r = \frac{4}{5} - 0 + 0 = \frac{4}{5}.$$ So, the series converges and its sum is $\frac{4}{5}$. 11. **Find $k$ such that** $$\sum_{r=1}^\infty (U_r + k U_{r+1}) = 1.$$ Using the sum formula, $$\sum_{r=1}^\infty U_r = \frac{4}{5},$$ and shifting index, $$\sum_{r=1}^\infty U_{r+1} = \sum_{m=2}^\infty U_m = \lim_{n \to \infty} \sum_{m=2}^n U_m = \lim_{n \to \infty} \left(\sum_{r=1}^n U_r - U_1\right) = \frac{4}{5} - U_1.$$ Calculate $U_1$: $$U_1 = \frac{4(2\cdot1 + 7)}{(2\cdot1 + 1)(2\cdot1 + 3)(2\cdot1 + 5)} = \frac{4(9)}{3 \cdot 5 \cdot 7} = \frac{36}{105} = \frac{12}{35}.$$ So, $$\sum_{r=1}^\infty U_{r+1} = \frac{4}{5} - \frac{12}{35} = \frac{28}{35} - \frac{12}{35} = \frac{16}{35}.$$ 12. **Set up equation:** $$\sum_{r=1}^\infty (U_r + k U_{r+1}) = \sum_{r=1}^\infty U_r + k \sum_{r=1}^\infty U_{r+1} = \frac{4}{5} + k \cdot \frac{16}{35} = 1.$$ Solve for $k$: $$k \cdot \frac{16}{35} = 1 - \frac{4}{5} = \frac{1}{5}.$$ Therefore, $$k = \frac{1/5}{16/35} = \frac{1}{5} \times \frac{35}{16} = \frac{7}{16}.$$ **Final answers:** - $A = 3$, $B = -1$ - $\sum_{r=1}^\infty U_r = \frac{4}{5}$ - $k = \frac{7}{16}$