Question

The term independent of $x$ in the expansion of ${\left[\frac{2\sqrt{x}}{5}-\frac{1}{2x\sqrt{x}}\right]}^{11}$ is

• A. $5^{\mathrm{th}}$ term
• B. $6^{\mathrm{th}}$ term
• C. ${11}^{\mathrm{th}}$ term
• D. No term

Answer 1

The correct option is D

No term

Explanation for the correct option:

Given expression is,

${\left[\frac{2\sqrt{x}}{5}-\frac{1}{2x\sqrt{x}}\right]}^{11}$

The binomial expansion of ${\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}$ is given as,

${\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}={}^{\mathrm{n}}\mathrm{C}_0{\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^0+{}^{\mathrm{n}}\mathrm{C}_1{\mathrm{a}}^{\mathrm{n}-1}{\mathrm{b}}^1+\dots +{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}{\mathrm{a}}^{\mathrm{n}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}}+\dots +{}^{\mathrm{n}}\mathrm{C}_{\mathrm{n}-1}{\mathrm{a}}^1{\mathrm{b}}^{\mathrm{n}-1}+{}^{\mathrm{n}}\mathrm{C}_{\mathrm{n}}{\mathrm{a}}^0{\mathrm{b}}^{\mathrm{n}}$

The general term of the expansion is given as,

${\mathrm{T}}_{\mathrm{r}+1}={}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}{\mathrm{a}}^{\mathrm{n}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}}$

Here, $\mathrm{n}=11$, $\mathrm{a}=\frac{2\sqrt{x}}{5}$ and $\mathrm{b}=-\frac{1}{2\mathrm{x}\sqrt{\mathrm{x}}}$

By finding the value of $\mathrm{r}$ at which the exponent of $x$ is $0$, we can find the term independent of $x$.

$\begin{array}{cc}\Rightarrow & {}^{11}\mathrm{C}_{\mathrm{r}}{\left(\frac{2\sqrt{x}}{5}\right)}^{11-\mathrm{r}}{\left(-\frac{1}{2\mathrm{x}\sqrt{\mathrm{x}}}\right)}^{\mathrm{r}}={\mathrm{kx}}^0\\ \Rightarrow & {}^{11}\mathrm{C}_{\mathrm{r}}·{\left(\frac{2}{5}\right)}^{11-\mathrm{r}}{\left(-\frac{1}{2}\right)}^{\mathrm{r}}{\left(\sqrt{x}\right)}^{11-\mathrm{r}}{\left(\frac{1}{\mathrm{x}\sqrt{\mathrm{x}}}\right)}^{\mathrm{r}}={\mathrm{kx}}^0\end{array}$

Where, $\mathrm{k}={}^{11}\mathrm{C}_{\mathrm{r}}·{\left(\frac{2}{5}\right)}^{11-\mathrm{r}}{\left(-\frac{1}{2}\right)}^{\mathrm{r}}$

Comparing the variables,

$\begin{array}{ccc}\Rightarrow & {\left(\sqrt{x}\right)}^{11-\mathrm{r}}{\left(\frac{1}{\mathrm{x}\sqrt{\mathrm{x}}}\right)}^{\mathrm{r}}=x^0& \\ \Rightarrow & {\left(x^{\frac{1}{2}}\right)}^{11-\mathrm{r}}{\left(\frac{1}{{\mathrm{x}}^{{\displaystyle \frac{3}{2}}}}\right)}^{\mathrm{r}}=x^0& \mathbf{.}\mathbf{.}\mathbf{.}\left[\because \sqrt{a}=a^{\frac{1}{2}},a\sqrt{a}=a^1\times a^{\frac{1}{2}}=a^{1+\frac{1}{2}}=a^{\frac{3}{2}}\right]\\ \Rightarrow & \left(x^{\frac{11-\mathrm{r}}{2}}\right)\left(x^{-\frac{3\mathrm{r}}{2}}\right)=x^0& ...\left[\because {\left(a^m\right)}^n=a^{mn},\frac{1}{a^m}=a^{-m}\right]\\ \Rightarrow & x^{\frac{11-\mathrm{r}-3\mathrm{r}}{2}}=x^0& ...\left[\because a^m\times a^n=a^{m+n}\right]\\ \Rightarrow & \frac{11-2-3\mathrm{r}}{2}=0& ...\left[\because a^m=a^n\Rightarrow m=n\right]\\ \Rightarrow & \mathrm{r}=\frac{11}{4}& \end{array}$

But, $\mathrm{r}$ cannot be a fraction.

Thus, no term independent of $x$.

Hence, option D is correct.