Question

The $11$th term in the expansion of ${(x+\frac{1}{\sqrt{x}})}^{14}$ is

• A. $\frac{999}{x}$
• B. $\frac{1001}{x}$
• C. $i$
• D. $\frac{x}{1001}$

Answer 1

The correct option is B $\frac{1001}{x}$

We know that $(r+1)$th term in the expansion of $(x+y{)}^n$ is given by ${}^n{C}_r{x}^{n-r}{y}^r$.

Hence $11$th term in the expansion of given given binomial is given by

${=}^{14}{C}_{10}{x}^4{\left(\frac{1}{\sqrt{x}}\right)}^{10}{=}^{14}{C}_{10}{x}^4\cdot \frac{1}{x^5}=\frac{{}^{14}{C}_{10}}{x}=\frac{1001}{x}$