If 1 x = 2 3! + 4 5! + 6 7! + , then find the value of: _ a 1^+ _a^x f(y) , f'(y) , _y(y^ x ) , dy given that f is a continuous, differentiable function whose graph passes through the point (1, 0)

Step-by-step solution

The given series is 1 x = 2 3! + 4 5! + 6 7! + The n^ th term of the series can be written as T_n = 2n (2n+1)! This can be simplified as T_n = (2n+1) - 1 (2n+1)! = 2n+1 (2n+1)! - 1 (2n+1)! = 1 (2n)! - 1 (2n+1)! Summing the series from n = 1 to gives: _ n=1 ^ T_n = ( 1 2! - 1 3! ) + ( 1 4! - 1 5! ) + ( 1 6! - 1 7! ) + 1 x = 1 2! - 1 3! + 1 4! - 1 5! + 1 6! - Using the expansion e^z = 1 + z + z^2 2! + z^3 3! + Substituting z = -1 , we get e⁻¹ = 1 - 1 + 1 2! - 1 3! + 1 4! - 1 5! + Thus, 1 x = e⁻¹ = 1 e , which implies