Question:

<p><strong>Step 1: Write the general term</strong></p><p>In the expansion of $\left(x^2 + \frac{2}{x}\right)^{15}$, the general term is:</p><p>$$T_{r+1} = \binom{15}{r}(x^2)^{15-r}\left(\frac{2}{x}\right)^r = \binom{15}{r}x^{2(15-r)} \cdot \frac{2^r}{x^r}$$</p><p>$$T_{r+1} = \binom{15}{r}2^r \cdot x^{30-2r-r} = \binom{15}{r}2^r \cdot x^{30-3r}$$</p><p><strong>Step 2: Find the term containing x^15</strong></p><p>For the coefficient of $x^{15}$, set the power equal to 15:</p><p>$$30 - 3r = 15$$</p><p>$$3r = 15$$</p><p>$$r = 5$$</p><p>Coefficient of $x^{15}$ is: $\binom{15}{5}2^5 = \binom{15}{5} \cdot 32$</p><p><strong>Step 3: Find the term independent of x</strong></p><p>For the term independent of x, set the power equal to 0:</p><p>$$30 - 3r = 0$$</p><p>$$r = 10$$</p><p>The constant term is: $\binom{15}{10}2^{10}$</p><p><strong>Step 4: Calculate the ratio</strong></p><p>$$\text{Ratio} = \frac{\binom{15}{5}2^5}{\binom{15}{10}2^{10}}$$</p><p>Since $\binom{15}{5} = \binom{15}{10}$ (symmetry property):</p><p>$$\text{Ratio} = \frac{2^5}{2^{10}} = \frac{1}{2^5} = \frac{1}{32}$$</p><p>Therefore, the ratio is $1 : 32$</p><p><strong>∴ Answer:</strong> B</p>