Smallest natural number n, such that coefficient of x in expansion of {left( {{x^2}, + ,frac{1

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7.Binomial Theorem

hard

The smallest natural number $n,$ such that the coefficient of $x$ in the expansion of ${\left( {{x^2}\, + \,\frac{1}{{{x^3}}}} \right)^n}$ is $^n{C_{23}}$ is

A

$38$

B

$58$

C

$23$

D

$35$

(JEE MAIN-2019)

Solution

${T_{r + 1}}\, = \,\sum\limits_{r = 0}^n {^n{C_r}\,{x^{2n – 2r}}\,.\,{x^{ – 3r}}} $

$2n – 5r = 1\Rightarrow 2n = 5r + 1$ for $r = 15, n = 38$ smallest value of $n$ is $38$.

Standard 11

Mathematics

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