Question

The term independent of $x$ in the expansion of ${\left[x^3-\frac{2}{x^2}\right]}^{15}$ is

• A. ${\mathrm{T}}_7$
• B. ${\mathrm{T}}_8$
• C. ${\mathrm{T}}_9$
• D. ${\mathrm{T}}_{10}$

Answer 1

The correct option is D

${\mathrm{T}}_{10}$

Explanation for the correct option:

Given expression is,

${\left[x^3-\frac{2}{x^2}\right]}^{15}$

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, $\mathrm{n}=15,\mathrm{a}={\mathrm{x}}^3$ and $\mathrm{b}=-\frac{2}{{\mathrm{x}}^2}$

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

$\begin{array}{cc}\Rightarrow & {}^{15}\mathrm{C}_{\mathrm{r}}{\left({\mathrm{x}}^3\right)}^{15}{\left(-\frac{2}{{\mathrm{x}}^2}\right)}^{\mathrm{r}}={\mathrm{kx}}^0\\ \Rightarrow & {}^{15}\mathrm{C}_{\mathrm{r}}·{\left(-2\right)}^{\mathrm{r}}{\left({\mathrm{x}}^3\right)}^{15}{\left(\frac{1}{{\mathrm{x}}^2}\right)}^{\mathrm{r}}={\mathrm{kx}}^0\end{array}$

Where, $\mathrm{k}={}^{15}\mathrm{C}_{\mathrm{r}}·{\left(-2\right)}^{\mathrm{r}}$

Comparing the variables,

$\begin{array}{ccc}\Rightarrow & {\left(x^3\right)}^{15-r}{\left(\frac{1}{x^2}\right)}^r=x^0& \\ \Rightarrow & \left(x^{45-3r}\right)\left(x^{-2r}\right)=x^0& ...\left[\because {\left(a^m\right)}^n=a^{mn},\frac{1}{a^m}=a^{-m}\right]\\ \Rightarrow & x^{45-3r-2r}=x^0& ...\left[\because a^m\times a^n=a^{m+n}\right]\\ \Rightarrow & 45-5r=0& ...\left[\because a^m=a^n\Rightarrow m=n\right]\\ \Rightarrow & r=9& \end{array}$

Thus, the term that is independent of $x$ is,

${\mathrm{T}}_{\mathrm{r}+1}={\mathrm{T}}_{9+1}={\mathrm{T}}_{10}$

Hence, option D is correct.