Find a Maclaurin series for f(x). (Usef* √ ₁ + 1³ d1f(x)=f(x) = x ++*12+Ën = 2(2n)!2"n! (2n-1)for 135... (2n-3).)(−1)²−¹(2n)!x5n+1( 5n + 1) (2n − 1) (2n!) ²

The objective of this question is to find the Maclaurin series for the given function f(x). The Maclaurin series is a type of Taylor series that is expanded about x = 0.The general form of a Maclaurin series is f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...In this case, the function f(x) is given as f(x) = x + sum from n=2 to 135 of ((-1)^(2n-1) * (2n)! * x^(5n+1)) / ((5n + 1) * (2n - 1) * (2n!)^2).The first term of the Maclaurin series is f(0), which is the value of the function at x = 0. For the given function, f(0) = 0.The second term of the Maclaurin series is f'(0)x, where f'(0) is the derivative of the function at x = 0. However, the derivative of the given function is complex and involves a sum with 135 terms. Therefore, it is not practical to compute the derivative directly.Instead, we can use the given formula to compute the terms of the Maclaurin series directly. The nth term of the Maclaurin series is given by ((-1)^(2n-1) * (2n)! * x^(5n+1)) / ((5n + 1) * (2n - 1) * (2n!)^2).We can compute the terms of the Maclaurin series for n = 2 to 135 using this formula. Note that the factorial function (2n)! and the power function x^(5n+1) can be computed using a scientific calculator or a computer program.Finally, the Maclaurin series for the given function is the sum of these terms, plus the first term f(0) = 0. Therefore, the Maclaurin series is f(x) = x + sum from n=2 to 135 of ((-1)^(2n-1) * (2n)! * x^(5n+1)) / ((5n + 1) * (2n - 1) * (2n!)^2).The Maclaurin series for the given function f(x) is obtained by computing the terms of the series using the given formula for n = 2 to 135. The series is the sum of these terms, plus the first term f(0) = 0.