KCET · Maths · Binomial Theorem
The 11th term in the expansion of \( \left(x+\frac{1}{\sqrt{x}}\right)^{14} \) is
- A \( \frac{999}{x} \)
- B \( \frac{1001}{x} \)
- C \( i \)
- D \( \frac{x}{1001} \)
Answer & Solution
Correct Answer
(B) \( \frac{1001}{x} \)
Step-by-step Solution
Detailed explanation
Given expression, \( \left(x+\frac{1}{\sqrt{x}}\right)^{14} \)
We know that nth term is given by
\[
\begin{array}{l}
T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r} \\
\text { So, } T_{11}={ }^{14} C_{10}\left(\chi^{4}\right)\left(\frac{1}{\sqrt{\chi}}\right)^{10} \\
={ }^{14} C_{10}\left(\chi^{4}\right)\left(\frac{1}{\chi^{5}}\right) \\
=\frac{{ }^{14} C_{10}}{\chi}=\frac{1001}{\chi}
\end{array}
\]
We know that nth term is given by
\[
\begin{array}{l}
T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r} \\
\text { So, } T_{11}={ }^{14} C_{10}\left(\chi^{4}\right)\left(\frac{1}{\sqrt{\chi}}\right)^{10} \\
={ }^{14} C_{10}\left(\chi^{4}\right)\left(\frac{1}{\chi^{5}}\right) \\
=\frac{{ }^{14} C_{10}}{\chi}=\frac{1001}{\chi}
\end{array}
\]
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