Question:

Step 1: Simplify the infinite series for $y$ by converting to a common base. We express each logarithm in terms of $\log_2 x$: $$y = \log_2 x + \log_4 x + \log_{16} x + \ldots$$ Using the change of base formula, $\log_{2^n} x = \frac{\log_2 x}{n}$, we get: $$y = \log_2 x + \frac{\log_2 x}{2} + \frac{\log_2 x}{4} + \ldots = \log_2 x\left(1 + \frac{1}{2} + \frac{1}{4} + \ldots\right)$$ This is a geometric series with first term 1 and common ratio $\frac{1}{2}$: $$y = \log_2 x \cdot \frac{1}{1-\frac{1}{2}} = 2\log_2 x$$ Step 2: Find the sum of the numerator $5+9+13+\ldots+(4y+1)$. This is an arithmetic progression with: - First term: $a = 5$ - Last term: $l = 4y+1$ - Common difference: $d = 4$ The number of terms is: $$n = \frac{4y+1-5}{4}+1 = \frac{4y-4}{4}+1 = y$$ The sum of an AP is: $$S_{\text{num}} = \frac{n(a+l)}{2} = \frac{y(5+4y+1)}{2} = \frac{y(4y+6)}{2} = y(2y+3)$$ Step 3: Find the sum of the denominator $1+3+5+\ldots+(2y-1)$. This is the sum of the first $y$ odd numbers, which equals $y^2$: $$S_{\text{den}} = y^2$$ Step 4: Simplify the right-hand side of the equation. $$\text{RHS} = \frac{y(2y+3)}{y^2} = \frac{2y+3}{y}$$ Step 5: Simplify the left-hand side of the equation. Convert $\log_4 x$ to base 2: $$4\log_4 x = 4 \cdot \frac{\log_2 x}{2} = 2\log_2 x$$ From Step 1, we know that $y = 2\log_2 x$, so: $$\text{LHS} = y$$ Step 6: Solve for $y$ by equating both sides. $$y = \frac{2y+3}{y}$$ Multiply both sides by $y$: $$y^2 = 2y+3$$ Rearrange to standard form: $$y^2 - 2y - 3 = 0$$ Factor: $$(y-3)(y+1) = 0$$ This gives $y = 3$ or $y = -1$. Since $y$ must be positive (as it represents a sum of logarithms), we have: $$y = 3$$ Step 7: Find the value of $x$. From $y = 2\log_2 x = 3$: $$\log_2 x = \frac{3}{2}$$ Therefore: $$x = 2^{3/2} = 2 \cdot 2^{1/2} = 2\sqrt{2}$$ Step 8: Calculate $x^2 y$. $$x^2 y = (2\sqrt{2})^2 \cdot 3 = (4 \cdot 2) \cdot 3 = 8 \cdot 3 = 24$$ **Final Answer:** The value of $x^2 y$ equals $\boxed{24}$, which corresponds to **Option 3**.