For m , n > 0 , let α m , n = ∫ 0 2 t m 1 + 3 t n d t . If , 11 α 10 , 6 + 18 α 11 , 5 = p 14 6 , then p is equal to

Step-by-step solution

Given, α m , n = ∫ 0 2 t m 1 + 3 t n d t Now using integration by parts we get, ⇒ α m , n = 1 + 3 t n ⋅ t m + 1 m + 1 0 2 − ∫ 0 2 n 1 + 3 t n − 1 × 3 ⋅ t m + 1 m + 1 d t ⇒ m + 1 α m , n = 1 + 3 t n ⋅ t m + 1 0 2 − 3 n m + 1 ∫ 0 2 1 + 3 t n − 1 × t m + 1 d t ⇒ m + 1 α m , n = 1 + 3 × 2 n ⋅ 2 m + 1 − 3 n m + 1 α m + 1 , n - 1 ⇒ m + 1 α m , n = 7 n · 2 m + 1 - 3 n α m +