Question

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:

• A. 10
• B. 8
• C. 14
• D. 7

Answer 1

Answer: (D)

7

Answer 2

The given integral is:
$I_n = \int_0^1 (1 - x^k)^n dx.$
Using integration by parts, we get:
$I_n = \frac{nk}{nk + 1} I_{n-1}.$
Iterating this formula, the relationship becomes:
$\frac{I_n}{I_{n-1}} = \frac{nk}{nk + 1}.$
Given:
$\frac{I_{21}}{I_{20}} = \frac{147}{148},$
we substitute into the formula:
$\frac{21k}{21k + 1} = \frac{147}{148}.$
Cross-multiplying and solving:
$148 \cdot 21k = 147 \cdot (21k + 1),$
$148 \cdot 21k = 147 \cdot 21k + 147,$
$21k = 147 \implies k = 7.$