Question

The coefficient of $x^{-10}$ in ${\left[x^2-\left(\frac{1}{x^3}\right)\right]}^{10}$is

• A. $-252$
• B. $210$
• C. $-(5!)$
• D. $-120$

Answer 1

The correct option is B

$210$

Explanation for the correct option:

Finding coefficient of $x^{-10}$ in given expression:

Given expression is ${\left[x^2-\left(\frac{1}{x^3}\right)\right]}^{10}$

General term of the expression is-

$$\begin{gathered} T_{r+1}={}^{10}C_r{\left(x^2\right)}^{10-r}{\left(\frac{-1}{x^3}\right)}^r \\ {T}_{r+1}={}^{10}C_r{\left(x\right)}^{20-2r}{(-1)}^r{\left(x\right)}^{-3r} \\ {T}_{r+1}={}^{10}C_r{\left(x\right)}^{20-5r}{(-1)}^r\dots \dots \left(1\right) \end{gathered}$$

Since, we need coefficient of $x^{-10}$,

$$\begin{gathered} \therefore 20-5r=-10 \\ \Rightarrow 5r=30 \\ \Rightarrow r=6 \end{gathered}$$

Substituting the value of $r$ in equation $\left(1\right)$ we get,

$$\begin{gathered} {}^{10}C_6{\left(x\right)}^{20-5\left(6\right)}{(-1)}^6 \\ ={}^{10}C_6{\left(x\right)}^{-10}{(-1)}^6 \end{gathered}$$

The coefficient of $x^{-10}$is

$$\begin{gathered} {}^{10}C_6{(-1)}^6=\frac{10!}{6!4!}\left[\because {}^{n}C_r=\frac{n!}{r!(n-1)!}\right] \\ =\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{\left(6\times 5\times 4\times 3\times 2\times 1\right)\left(4\times 3\times 2\times 1\right)} \\ {}^{10}C_6{(-1)}^6=210 \end{gathered}$$

Hence, Option(B) is correct.