91Ó°ÊÓ

Since the power series has a term x^(2k+1), we can replace (2k+1) with a common power 'n' to make the series look like a common Maclaurin series representation. $$\sum_{k=0}^{\infty} 2^{k} x^{2 k+1} = \sum_{n=1}^{\infty} a_n x^{n}$$ with n = 2k+1 and hence k = (n-1)/2.

From the above equation and replacing k = (n-1)/2 we get, $$a_n = 2^{(n-1)/2}$$

We need to recognize the above power series as the Maclaurin series (Taylor series around x=0) of a known function. The Maclaurin series for the function g(x) is given by: $$g(x) =\sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n!}x^n$$ where g^(n)(0) is the n-th derivative of g(x) evaluated at x = 0. Our goal is to find such a function g(x) using the coefficients we've found before.

We can compare our power series with the known Maclaurin series of the function \(\frac{1}{(1-x)^{\alpha}}\), which is given as: $$\frac{1}{(1-x)^{\alpha}} = \sum_{n=0}^{\infty} \binom{\alpha + n - 1}{n} x^n$$ For our problem, we can simplify the known function's power series to match our power series $$\sum_{n=1}^{\infty} \binom{\alpha + n - 1}{n} x^n = \sum_{n=1}^{\infty} a_n x^{n}$$ $$\binom{\alpha + n - 1}{n} = 2^{(n-1)/2}$$ Observing the coefficients, we notice that for odd values of n (which is 2k+1), the binomial coefficients become zero for the even powers of x (which don't exist in this power series). To achieve this, we want to choose the parameter α such that the power series only contains odd powers of x. By choosing α = -1/2, it turns the binomial coefficient into simple 2^(n-1)/2. Therefore, the function that we are looking for is: $$g(x) = \frac{1}{(1-x)^{-\frac{1}{2}}}$$

Finally, we found that our given power series is the Maclaurin series representation of the function given by: $$g(x) = \frac{1}{(1-x)^{-\frac{1}{2}}}$$