Discussion
The constant term in \(\left(\frac{\sqrt{x}}{2} + \frac{1}{3x^{2}}\right)^{10}\) is
Answer: C
The constant term is \(\frac{5}{256}\).
The general term is
\[ \begin{aligned} T_{r+1} &= {}^{10}C_{r}\left(\frac{\sqrt{x}}{2}\right)^{10-r} \\ &\quad \times \left(\frac{1}{3x^{2}}\right)^{r} \end{aligned} \]
with power of x equal to \(\frac{10 - r}{2} - 2r\).
\[ \begin{aligned} \frac{10 - r}{2} - 2r = 0 &\Rightarrow r = 2 \\ T_{3} = {}^{10}C_{2}\cdot\frac{1}{2^{8}}\cdot\frac{1}{3^{2}} &= \frac{45}{256 \times 9} = \frac{5}{256} \end{aligned} \]
Using \(2^{7} = 128\) instead of \(2^{8}\) gives 5/128, and forgetting to square the 3 leaves a 9 or 15 in the wrong place. The term exists, so 0 is wrong.
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