The smallest natural number n , such that the coefficient of x in...

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

C23 n$ , is

Options

A $58$
B $38$
C $35$
D $23$

Correct answer

B. $38$

Step-by-step solution

In the expansion of ${\left({x}^{2} + \dfrac{1}{{x}^{3}}\right)}^{n} the general term is$ {T}_{r + 1} = {C}_{r} n {\left({x}^{2}\right)}^{n - r}{\left(\dfrac{1}{{x}^{3}}\right)}^{r} $$ = {C}_{r} n {x}^{2n - 2r - 3r} = {C}_{r} n {x}^{2n - 5r} $For coefficient of$ x $,$ 2n - 5r = 1 $$ ⇒ r = \dfrac{2n - 1}{5} $So, we have the coefficient as$ C\dfrac{2n - 1}{5} n $Using, the given value and$ Cr n = Cn - r n $$ ⇒ C\dfrac{2n - 1}{5} n = {C}_{23} = {C}_{n - 23} n n $If$ \dfrac{2n - 1}{5} = 23 ⇒ n = 58 and if $\dfrac{2n - 1}{5} = n - 23 ⇒ n = 38$

Thus, the minimum value of $'n'$ is $38 .$

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