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

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:

& {{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{aligned}$$ Now we have with us the coefficient of x when we expand the binomial as $^{n}{{C}_{23}}$ given in the question. Since the power of x is 1, we will compare the ${{\left( r+1 \right)}^{th}}$ term with it: $${{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}}$$ As coefficient of x is $^{n}{{C}_{23}}$, therefore either $r=23$ or $n-r=23$ Let $r=23$, putting this value in equation1 we will get: $\begin{aligned} & 2n=5r+1\Rightarrow 2n=5\left( 23 \right)+1\Rightarrow 2n=116 \\\ & \Rightarrow n=58 \\\ \end{aligned}$ Now let $n-r=23$ , then $r=n-23$ putting this value in equation 1: $\begin{aligned} & 2n=5r+1\Rightarrow 2n=5\left( n-23 \right)+1\Rightarrow 2n=5n-115+1 \\\ & \Rightarrow 3n=114 \\\ & \Rightarrow n=38 \\\ \end{aligned}$ Since we require the smallest natural number $n$ we will have our answer: $n=38$ **Hence, option B is correct.** **Note** : The total number of terms in the binomial expansion of ${{\left( a+b \right)}^{n}}$ is (n+1), that is one more than the exponent n. Do not forget to take both the condition $$^{n}{{C}_{r}}$$ and $$^{n}{{C}_{n-r}}$$, by doing this we will see what value will be our answer according to the question.