1. **State the problem:** Calculate the value of $$A = \sqrt{\frac{5 \cdot 4^{n-1}}{2^{2n-2} + 4^{n-2}}}$$ for some integer $n$. 2. **Rewrite the expression using properties of exponents:** Recall that $4 = 2^2$, so we can rewrite powers of 4 in terms of 2: $$4^{n-1} = (2^2)^{n-1} = 2^{2(n-1)} = 2^{2n-2}$$ $$4^{n-2} = (2^2)^{n-2} = 2^{2(n-2)} = 2^{2n-4}$$ 3. **Substitute these back into the expression:** $$A = \sqrt{\frac{5 \cdot 2^{2n-2}}{2^{2n-2} + 2^{2n-4}}}$$ 4. **Factor the denominator:** Factor out the smaller power $2^{2n-4}$: $$2^{2n-2} + 2^{2n-4} = 2^{2n-4}(2^2 + 1) = 2^{2n-4} \cdot 5$$ 5. **Rewrite $A$ with the factored denominator:** $$A = \sqrt{\frac{5 \cdot 2^{2n-2}}{5 \cdot 2^{2n-4}}}$$ 6. **Cancel the common factor 5:** $$A = \sqrt{\frac{\cancel{5} \cdot 2^{2n-2}}{\cancel{5} \cdot 2^{2n-4}}}$$ 7. **Simplify the powers of 2 inside the square root:** $$A = \sqrt{2^{(2n-2) - (2n-4)}} = \sqrt{2^{2}}$$ 8. **Evaluate the square root:** $$A = \sqrt{2^2} = 2$$ **Final answer:** $A = 2$ This corresponds to option ⓑ 2.