Question:
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 :
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 :
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Write the general term, equate the power of x to 1 and use \(^{n}\mathrm{C}_r={}^{n}\mathrm{C}_{n-r}\) to get two possible values of n.
Updated On: Nov 5, 2026
- 23
- 58
- 38
- 35
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The Correct Option is C
Solution and Explanation
General term: \(T_{r+1}={}^{n}\mathrm{C}_r (x^2)^{n-r}\left(\frac{1}{x^3}\right)^r={}^{n}\mathrm{C}_r\,x^{2n-5r}\).
Coefficient of x: \(2n-5r=1\), so the coefficient is \(^{n}\mathrm{C}_r\).
Given \(^{n}\mathrm{C}_r={}^{n}\mathrm{C}_{23}\), so either \(r=23\) or \(r=n-23\).
Case 1: \(r=23\): \(2n-115=1\Rightarrow n=58\).
Case 2: \(r=n-23\): \(2n-5(n-23)=1\Rightarrow -3n+115=1\Rightarrow n=38\).
The smallest natural number is \(n=38\).
Answer: (3) 38
Coefficient of x: \(2n-5r=1\), so the coefficient is \(^{n}\mathrm{C}_r\).
Given \(^{n}\mathrm{C}_r={}^{n}\mathrm{C}_{23}\), so either \(r=23\) or \(r=n-23\).
Case 1: \(r=23\): \(2n-115=1\Rightarrow n=58\).
Case 2: \(r=n-23\): \(2n-5(n-23)=1\Rightarrow -3n+115=1\Rightarrow n=38\).
The smallest natural number is \(n=38\).
Answer: (3) 38
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