Let for x ∈ ℝ , S 0 x = x , S k x = C k x + k ∫ 0 x S k - 1 t d t where C 0 = 1 , C k = 1 - ∫ 0 1 S k - 1 x d x , k = 1 , 2 , 3 , … Then S 2 3 + 6 C 3 is equal to _ _ _ _ _ _ _ .

Step-by-step solution

Given, S 0 x = x ,   S k x = C k x + k ∫ 0 x S k - 1 t d t Now for S 0 x = x , C 0 = 1 Now solving, C k = 1 - ∫ 0 1 S k - 1 x d x for k = 1 we get, C 1 = 1 - ∫ 0 1 x d x = 1 2 Now putting k = 1 in S k x = C k x + k ∫ 0 x S k - 1 t d t we get, ⇒ S 1 x = x 2 + 1 · ∫ 0 x t d t = x 2 + x 2 2 Now for k = 2 we get, C 2 = 1 - ∫ 0 1 x 2 + x 2 2 d x = 7 12 And S 2 x = 7 12 x + 2 ∫ 0 x t 2 + t 2 2 d t ⇒ S 2 x = 7 x 12 + x 2 2 + x 3 3 Now taking k = 3 we get, C 3