Question

The natural number $m$, for which the coefficient of $x$ in the binomial expansion of ${\left(x^m+\frac{1}{x^2}\right)}^{22}$ is $1540$ is

Answer 1

Step 1: Find the general term

Given: The coefficient of $x$ in the binomial expansion of ${\left(x^m+\frac{1}{x^2}\right)}^{22}$ is $1540$.

We know that for a binomial expansion of ${\left(a+b\right)}^n$, $T_{r+1}={}^{n}C_r{a}^{n-r}{b}^r$.

So for the binomial expansion of ${\left(x^m+\frac{1}{x^2}\right)}^{22}$we have,

$$\begin{gathered} T_{r+1}={}^{22}C_r{\left(x^m\right)}^{22-r}{\left(\frac{1}{x^2}\right)}^r\left[\because a=x^m,b=\frac{1}{x^2}\&n=22\right] \\ ={}^{22}C_r{\left(x\right)}^{22m-mr}{\left(x\right)}^{-2r} \\ ={}^{22}C_r{\left(x\right)}^{22m-mr-2r} \end{gathered}$$

Step 2: Find the value of ‘m’

As we have given that the coefficient of $x$ is $1540$.

So, we can say,

For $22m-mr-2r=1$, ${}^{22}C_r=1540$.

We know that,

$$\begin{gathered} {}^{22}C_3=\frac{22\times 21\times 20}{3\times 2\times 1} \\ =1540 \end{gathered}$$

And ${}^{22}C_3={}^{22}C_{22-3}={}^{22}C_{19}\left[\because {}^{n}C_r={}^{n}C_{n-r}\right]$

So, we have two possible values of $r$ as $3$ and $19$

Now, for $r=3$

$$\begin{gathered} 22m-mr-2r=1 \\ \Rightarrow 22m-3m-2\left(3\right)=1 \\ \Rightarrow 19m-6=1 \\ \Rightarrow 19m=7 \end{gathered}$$

$\Rightarrow m=\frac{7}{19}$, which is not possible as $m$is a natural number.

So, for $r=19$

$$\begin{gathered} 22m-mr-2r=1 \\ \Rightarrow 22m-19m-2\left(19\right)=1 \\ \Rightarrow 3m-38=1 \\ \Rightarrow 3m=39 \\ \Rightarrow m=\frac{39}{3} \\ \Rightarrow m=13\in \mathrm{\mathbb{N}} \end{gathered}$$

Hence, the required value of $m$is $3$.