Question

The smallest natural number$n$ , such that the coefficient of $x$ in the expansion of ${{\left( {{x}^{2}}+\dfrac{1}{{{x}^{3}}} \right)}^{n}}$ is $^{n}{{C}_{23}}$ is:
A. $35$
B.$38$
C. $23$
D. $58$

Answer 1

Hint: e will first write the general term or (r+1) th term of the given binomial then we will compare the power of x obtained to 1. We will get the equation in n and r and then we will again use the coefficient $^{n}{{C}_{23}}$ and get the values of r. Finally we will get two values of r and we will choose the smaller one as asked in the question.

Complete step-by-step answer:
Now we know that according to Binomial theorem:
 If a and b are real numbers and n is a positive integer, then \[{{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+.............{{+}^{n}}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+{{.......}^{n}}{{C}_{n}}{{b}^{n}}\] , where \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\text{ where }0\le r\le n\] .
The general term or ${{\left( r+1 \right)}^{th}}$ term in the expansion is given by \[{{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]
We are given the binomial as ${{\left( {{x}^{2}}+\dfrac{1}{{{x}^{3}}} \right)}^{n}}$ and we are given that the coefficient of x when we expand the binomial is $^{n}{{C}_{23}}$
Now, first of all we will write its ${{\left( r+1 \right)}^{th}}$ term of the given binomial, that is:
\[\begin{align}
  & {{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{\left( {{x}^{2}} \right)}^{n-r}}{{\left( \dfrac{1}{{{x}^{3}}} \right)}^{r}} \\
 & \Rightarrow {{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{2n-2r}}\dfrac{1}{{{x}^{3r}}}\Rightarrow {{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{2n-2r}}{{x}^{-3r}}\Rightarrow {{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{2n-2r+\left( -3r \right)}} \\
 & \Rightarrow {{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{2n-5r}} \\
\end{align}\]
Now we have with us the coefficient of x when we expand the binomial as $^{n}{{C}_{23}}$ given in the question. Since the power of x is 1, we will compare the ${{\left( r+1 \right)}^{th}}$ term with it:
\[{{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{2n-5r}}\] Here we will equate the power of x that is $2n-5r$ to $1$.
$\therefore 2n-5r=1\Rightarrow 2n=5r+1\text{ }.........\text{Equation 1}\text{.}$
Now we know that \[^{n}{{C}_{r}}\] can be written as \[^{n}{{C}_{n-r}}\]
As coefficient of x is $^{n}{{C}_{23}}$, therefore either $r=23$ or $n-r=23$

Let $r=23$, putting this value in equation1 we will get:
$\begin{align}
  & 2n=5r+1\Rightarrow 2n=5\left( 23 \right)+1\Rightarrow 2n=116 \\
 & \Rightarrow n=58 \\
\end{align}$
Now let $n-r=23$ , then $r=n-23$ putting this value in equation 1:
$\begin{align}
  & 2n=5r+1\Rightarrow 2n=5\left( n-23 \right)+1\Rightarrow 2n=5n-115+1 \\
 & \Rightarrow 3n=114 \\
 & \Rightarrow n=38 \\
\end{align}$
Since we require the smallest natural number $n$ we will have our answer: $n=38$
Hence, option B is correct.

Note: The total number of terms in the binomial expansion of ${{\left( a+b \right)}^{n}}$ is (n+1), that is one more than the exponent n. Do not forget to take both the condition \[^{n}{{C}_{r}}\] and \[^{n}{{C}_{n-r}}\], by doing this we will see what value will be our answer according to the question.