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 ${ }^{n} \mathrm{C}_{23}$ is

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Binomial Theorem

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

(x^{2} + \frac{1}{x ^{3}})^{n}

^{n} C_{23}

Answer: (A)

The general term of

(x^{2} + \frac{1}{x ^{3}})^{n}

(r n) x^{2 (n - r) - 3 r} = (r n) x^{2 n - 5 r}

. For the power of

x

to be

1

we need

2 n - 5 r = 1

. The coefficient is given as

(23 n)

, and since

(r n) = (23 n)

allows either

r = 23

r = n - 23

: the first gives

n = 58

, while

r = n - 23

turns

2 n - 5 (n - 23) = 1

into

n = 38

. The smallest natural number is

38

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option