QJEEM19Ma2691: JEE Main 2019 Mathematics - Class 11 JEEM19Ma2691 Solution: The smallest natural number n, such that the coefficient of x…

Question

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

{\left( {{x^2} + {1 \over {{x^3}}}} \right)^n}

is ⁿC₂₃, is :

Accepted Answer

General term

{T_{r + 1}} = {}^n{C_r}{x^{2n - 2r}}.{x^{ - 3r}}

\therefore

2n - 5r = 1 \Rightarrow 2n = 5r + 1

\therefore

r = {{2n - 1} \over 5}

\Rightarrow

Coefficient of x =

{}^n{C_{\left( {{{2n - 1} \over 5}} \right)}} = {}^n{C_{23}}

\Rightarrow

{{2n - 1} \over 5} = 23\,\,or\,\,n - \left( {{{2n - 1} \over 5}} \right) = 23

\Rightarrow

2n - 1 = 115

\Rightarrow

n = 58

and n = 38

\therefore

smallest n = 38

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