Question:medium

The value of $k \in \mathbb{N}$ for which the integral \[ I_n = \int_0^1 (1 - x^k)^n \, dx, \, n \in \mathbb{N}, \] satisfies $147 \, I_{20} = 148 \, I_{21}$ is:

Updated On: Jan 13, 2026
  • 10
  • 8
  • 14
  • 7
Show Solution

The Correct Option is D

Solution and Explanation

\[ I_n = \int_{0}^{1} (1 - x^k)^n \cdot 1\, dx \] \[ I_n = (1 - x^k)^n \cdot x - nk \int_{0}^{1} (1 - x^k)^{n-1} \cdot x^{k-1}\, dx \] \[ I_n = nk \left[ \int_{0}^{1} \left( (1 - x^k)^n - (1 - x^k)^{n-1} \right) dx \right] \] \[ I_n = nk I_n - nk I_{n-1} \] \[ \frac{I_n}{I_{n-1}} = \frac{nk}{nk + 1} \] \[ \frac{I_{21}}{I_{20}} = \frac{21k}{1 + 21k} \] \[ = \frac{147}{148} \Rightarrow k = 7 \]

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