Question

Let f_n = \(\int\limits_0^{\pi/2}\) \(\left(\displaystyle\sum_{k=1}^{n}\,sin^{k-1}x\right)\left(\displaystyle\sum_{k=1}^{n}\,(2k-1)sin^{k-1}x\right)\) cosx dx, n ∈ N. Then f₂₁ – f₂₀ is equal to____.

Answer 1

Correct Answer: 41

We are given the following expression for \( f_n \): \[ f_n = \int_0^{\frac{\pi}{2}} \left( \sum_{k=1}^{n} \sin^{k-1} x \right) \left( \sum_{k=1}^{n} (2k-1) \sin^{k-1} x \right) \cos x \, dx. \] Let \( \sin x = t \), so that \( \cos x \, dx = dt \), and the integral becomes: \[ f_n = \int_0^1 \left( \sum_{k=1}^{n} t^{k-1} \right) \left( \sum_{k=1}^{n} (2k-1) t^{k-1} \right) dt. \] Now, expand the sums and compute the integrals for each case. For \( f_{n+1} - f_n \), we get: \[ f_{n+1} - f_n = \int_0^1 \left( 1 + t + t^2 + \cdots + t^n \right) \left( 1 + 3t + 5t^2 + \cdots + (2n+1)t^n \right) dt. \] Putting \( n = 20 \), we find: \[ f_{21} - f_{20} = \int_0^1 (1 + 3t + 5t^2 + \cdots + 41t^{20}) dt. \] Simplify the expression: \[ f_{21} - f_{20} = \left( \frac{1}{21} + \frac{3}{22} + \frac{5}{23} + \cdots + \frac{41}{40} \right). \] The sum simplifies to \( 40 + 1 = 41 \).