Let f n = ∫ 0 π 2 ∑ k = 1 n sin k - 1 x ∑ k = 1 n ( 2 k - 1 ) sin k - 1 x cos x d x , n ∈ ℕ . Then f 21 - f 20 is equal to

Step-by-step solution

Given, f n = ∫ 0 π / 2 ∑ k = 1 n sin k - 1 x ∑ k = 1 n ( 2 k - 1 ) sin k - 1 x cos x d x Now let, sin x = t ⇒ cos x d x = d t So, f n = ∫ 0 1 ∑ k = 1 n ( t ) k - 1 ∑ k = 1 n ( 2 k - 1 ) ( t ) k - 1 d t ⇒ f n = ∫ 0 1 1 + t + t 2 . . . . + t n - 1 1 + 3 t + 5 t 2 + . . . . . + 2 n - 1 t n - 1 d t Now multiply and divide by t we get, ⇒ f n = ∫ 0 1 t 1 2 + t 3 2 + t 5 2 . . . . + t 2 n - 1 2 t 1 + 3 t + 5 t 2 + . . . . . + 2 n - 1 t n - 1 d t &#8658