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General Term of Binomial Expansion

The smallest ...

Question

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

23

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38

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35

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58

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Solution

The correct option is B $38$
Given,
${(x^{2} + \frac{1}{x^{3}})}^{n}$ , its $(r + 1)^{t h}$ term, is

$T_{r + 1} =^{n} C_{r} x^{2 n - 2 r} \cdot x^{- 3 r}$
$=^{n} C_{r} x^{2 n - 5 r}$

$∴ 2 n - 5 r = 1 \Rightarrow r = \frac{2 n - 1}{5}$

$∴ Coeff. of x =^{n} C_{(\frac{2 n - 1}{5})} =^{n} C_{23}$
$∴ \frac{2 n - 1}{5} = 23$ or $n - (\frac{2 n - 1}{5}) = 23$
$\Rightarrow n = 58$ or $n = 38$

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