JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 20)
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 :
23
58
38
35
Explanation
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
$${T_{r + 1}} = {}^n{C_r}{x^{2n - 2r}}.{x^{ - 3r}}$$
$$ \therefore $$ $$2n - 5r = 1 \Rightarrow 2n = 5r + 1$$
$$ \therefore $$ $$r = {{2n - 1} \over 5}$$
$$ \Rightarrow $$ 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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