Question

If 
\(n(2n+1)\int_{0}^{1}(1−xn)^{2n}dx=1177\int_{0}^{1}(1−x^n)^{2n+1}dx\)
,then n ∈ N is equal to ______.

Answer 1

Correct Answer: 24

The correct answer is 24
\(\int_{0}^{1} (1 - x^n)^{2n+1} \,dx = \int_{0}^{1} (1 - x^n)^{2n+1} \,dx\)
\(= \left[ (1 - x^n)^{2n+1} \cdot x \right]_{0}^{1} - \int_{0}^{1} x \cdot (2n+1)(1 - x^n)^{2n} \cdot (-nx^{n-1}) \,dx\)
\(= n(2n+1) \int_{0}^{1} (1 - (1 - x^n))(1 - x^n)^{2n} \,dx\)
\(= n(2n+1) \int_{0}^{1} (1 - x^n)^{2n} \,dx - n(2n+1) \int_{0}^{1} (1 - x^n)^{2n+1} \,dx\)
\((1 + n(2n+1)) \int_{0}^{1} (1 - x^n)^{2n+1} \,dx = n(2n+1) \int_{0}^{1} (1 - x^n)^{2n} \,dx\)
\((2n^2 + n + 1) \int_{0}^{1} (1 - x^n)^{2n+1} \,dx = 1177\int_{0}^{1} (1 - x^n)^{2n+1} \,dx\)
∴ 2n² + n + 1 = 1177
2n² + n – 1176 = 0
\(\therefore n=24 or −\frac{49}{2}\)
∴ n = 24