Discussion
The coefficient of \(x^{5}\) in the binomial expansion of \(\left(\sqrt{x} + \frac{1}{x}\right)^{10}\) is
Answer: E
The coefficient of \(x^{5}\) is 1.
The general term of \(\left(\sqrt{x} + \frac{1}{x}\right)^{10}\) is
\[ \begin{aligned} T_{r+1} &= \binom{10}{r} (x^{1/2})^{10-r} (x^{-1})^{r} \\ &= \binom{10}{r} x^{(10 - 3r)/2} \end{aligned} \]
For \(x^{5}\), \(\frac{10 - 3r}{2} = 5\), so r = 0. The term is the very first one, \(\binom{10}{0}(\sqrt{x})^{10} = x^{5}\), whose coefficient is 1.
The other numbers are coefficients of other terms: 10 and 45 are \(\binom{10}{1}\) and \(\binom{10}{2}\), the coefficients of \(x^{7/2}\) and \(x^{2}\).
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