1. **State the problem:** We want to verify or solve the equation: $$\frac{3}{1 \cdot 2 \cdot 2} + \frac{4}{2 \cdot 3 \cdot 2^3} + \frac{5}{3 \cdot 4 \cdot 2^3} + \frac{n + 2}{n (n + 1) 2^n} = 1 - \frac{1}{(n + 1) 2^n}$$ 2. **Understand the terms:** Each term on the left is a fraction with products in the denominator involving consecutive integers and powers of 2. 3. **Simplify the first three terms:** - First term: $$\frac{3}{1 \cdot 2 \cdot 2} = \frac{3}{4}$$ - Second term: $$\frac{4}{2 \cdot 3 \cdot 2^3} = \frac{4}{2 \cdot 3 \cdot 8} = \frac{4}{48} = \frac{1}{12}$$ - Third term: $$\frac{5}{3 \cdot 4 \cdot 2^3} = \frac{5}{3 \cdot 4 \cdot 8} = \frac{5}{96}$$ 4. **Sum the first three terms:** Find common denominator for $$\frac{3}{4}, \frac{1}{12}, \frac{5}{96}$$ which is 96: $$\frac{3}{4} = \frac{72}{96}, \quad \frac{1}{12} = \frac{8}{96}, \quad \frac{5}{96} = \frac{5}{96}$$ Sum: $$\frac{72}{96} + \frac{8}{96} + \frac{5}{96} = \frac{85}{96}$$ 5. **Rewrite the equation with sum of first three terms:** $$\frac{85}{96} + \frac{n + 2}{n (n + 1) 2^n} = 1 - \frac{1}{(n + 1) 2^n}$$ 6. **Isolate the term with $n$:** Move $$\frac{85}{96}$$ to the right: $$\frac{n + 2}{n (n + 1) 2^n} = 1 - \frac{1}{(n + 1) 2^n} - \frac{85}{96}$$ 7. **Simplify the right side:** Calculate $$1 - \frac{85}{96} = \frac{96}{96} - \frac{85}{96} = \frac{11}{96}$$ So: $$\frac{n + 2}{n (n + 1) 2^n} = \frac{11}{96} - \frac{1}{(n + 1) 2^n}$$ 8. **Find common denominator on right side:** Denominator is $$96 (n + 1) 2^n$$ Rewrite terms: $$\frac{11}{96} = \frac{11 (n + 1) 2^n}{96 (n + 1) 2^n}$$ $$\frac{1}{(n + 1) 2^n} = \frac{96}{96 (n + 1) 2^n}$$ So right side: $$\frac{11 (n + 1) 2^n - 96}{96 (n + 1) 2^n}$$ 9. **Set equal to left side:** $$\frac{n + 2}{n (n + 1) 2^n} = \frac{11 (n + 1) 2^n - 96}{96 (n + 1) 2^n}$$ 10. **Cross multiply:** $$ (n + 2) \cdot 96 (n + 1) 2^n = n (n + 1) 2^n \cdot \left(11 (n + 1) 2^n - 96\right) $$ 11. **Simplify both sides:** Left: $$96 (n + 2)(n + 1) 2^n$$ Right: $$n (n + 1) 2^n \left(11 (n + 1) 2^n - 96\right) = n (n + 1) 2^n \cdot 11 (n + 1) 2^n - n (n + 1) 2^n \cdot 96$$ $$= 11 n (n + 1)^2 2^{2n} - 96 n (n + 1) 2^n$$ 12. **Divide both sides by $$ (n + 1) 2^n $$ (assuming $$n \neq -1$$):** Left: $$96 (n + 2)$$ Right: $$11 n (n + 1) 2^n - 96 n$$ 13. **Rewrite equation:** $$96 (n + 2) = 11 n (n + 1) 2^n - 96 n$$ 14. **Bring all terms to one side:** $$11 n (n + 1) 2^n - 96 n - 96 (n + 2) = 0$$ 15. **Expand:** $$11 n (n + 1) 2^n - 96 n - 96 n - 192 = 0$$ $$11 n (n + 1) 2^n - 192 n - 192 = 0$$ 16. **Rewrite:** $$11 n (n + 1) 2^n = 192 n + 192$$ 17. **Factor right side:** $$192 (n + 1)$$ 18. **Divide both sides by $$ (n + 1) $$ (assuming $$n \neq -1$$):** $$11 n 2^n = 192$$ 19. **Solve for $$n$$:** We want integer or real $$n$$ such that: $$11 n 2^n = 192$$ 20. **Check integer values:** - For $$n=3$$: $$11 \times 3 \times 2^3 = 11 \times 3 \times 8 = 264 \neq 192$$ - For $$n=2$$: $$11 \times 2 \times 4 = 88 \neq 192$$ - For $$n=4$$: $$11 \times 4 \times 16 = 704 \neq 192$$ 21. **Approximate solution:** Rewrite as: $$n 2^n = \frac{192}{11} \approx 17.4545$$ Try $$n=3.0$$ gives 24, too high; $$n=2.5$$ gives about $$2.5 \times 5.656 = 14.14$$ too low. So $$n \approx 2.7$$ approximately. **Final answer:** The equation holds for $$n$$ approximately satisfying: $$11 n 2^n = 192$$ or $$n 2^n = \frac{192}{11}$$ This is the implicit solution for $$n$$. --- **Summary:** The given sum equals the right side if and only if $$n$$ satisfies $$11 n 2^n = 192$$.