Question

If the constant term in the binomial expansion of $(\sqrt{x} - \frac{k}{x^2})^{10}$ is 405, then |k| equals:

• A. 2
• B. 3
• C. 9
• D. 1

Answer 1

Answer: (B)

3

Answer 2

The general term of $(\sqrt{x} - \frac{k}{x^2})^{10}$ is $T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (-kx^{-2})^r$. The power of $x$ is $\frac{10-r}{2} - 2r$. For the constant term, this power is 0, which gives $r=2$. The constant term is $\binom{10}{2}(-k)^2 = 45k^2$. Given $45k^2 = 405$, we find $k^2=9$. Therefore, $|k|=3$.