WAEC Mathematics 2014 Paper

Concept:Express every number in the expression as a power of $2$ and apply the laws of indices.Explanation:Write $8 = 2^3$, $4 = 2^2$, and $16 = 2^4$.Then the expression becomes:$$\frac{\sqrt{(2^3)^2 \times (2^2)^{n+1}}}{2^{2n} \times 2^4}$$Simplify the powers inside the numerator:$$(2^3)^2 = 2^6 \quad \text{and} \quad (2^2)^{n+1} = 2^{2n+2}$$So the numerator under the square root becomes:$$2^6 \times 2^{2n+2} = 2^{2n+8}$$Now take the square root:$$\sqrt{2^{2n+8}} = 2^{\frac{2n+8}{2}} = 2^{n+4}$$The denominator is:$$2^{2n} \times 16 = 2^{2n} \times 2^4 = 2^{2n+4}$$Therefore:$$\frac{2^{n+4}}{2^{2n+4}} = 2^{n+4 - (2n+4)} = 2^{-n}$$So the simplified value is $2^{-n}$, which does not match any of the given options.Answer:$2^{-n}$; none of the listed options is correct.