If 1 + x n = C 0 + C 1 x + C 2 x 2 + . … … . + C n x n , ∀ n ∈ N and C 0 2 1 + C 1 2 2 + C 2 2 3 + . … … + C n 2 n + 1 = λ 2 n + 1 ! n + 1 ! 2 , then the value of λ is equal to

Step-by-step solution

Consider 1 + x n = C 0 + C 1 x + C 2 x 2 + . … … . + C n x n . ∴ ∫ 0 x 1 + x n d x = C 0 x + C 1 x 2 2 + C 2 x 2 3 + . … … . + C n x n + 1 n + 1 ⇒ 1 + x n + 1 - 1 n + 1 = C 0 x + C 1 x 2 2 + C 2 x 3 3 + . … … . + C n x n + 1 n + 1 Also, x + 1 n = C 0 x n + C 1 x n - 1 + C 2 x n - 2 + . … … + C n Multiply and compare x n + 1 C 0 2 1 + C 1 2 2 + C 2 2 3 + . … … + C n 2 n + 1 = Coefficient of x n + 1 in x + 1 n 1 + x n + 1 - 1 n + 1 = x + 1