Define generating function for a sequence \( a_{0}, a_{1}, a_{2}, \ldots \) Use generating function method to solve the recurrence relatio \[ 2 a_{n}=a_{n-1}+2^{n} \] For all \( n \geq 1 \) with \( a_{0}=1 \).

A generating function for a sequence \( a_{0}, a_{1}, a_{2}, \ldots \) is defined as the formal power series \[ A(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + \ldots = \sum_{n=0}^{\infty} a_{n} x^{n}. \] To solve the recurrence relation \( 2 a_{n} = a_{n-1} + 2^{n} \) using generating functions, we can start by multiplying both sides of the equation by \( x^{n} \) and summing for \( n \geq 1 \): \[ \sum_{n=1}^{\infty} 2a_{n}x^{n} = \sum_{n=1}^{\infty} a_{n-1}x^{n} + \sum_{n=1}^{\infty} 2^{n}x^{n}. \] The left-hand side becomes \( 2\left(A(x) - a_{0}\right) \). The right-hand side can be rewritten as \( xA(x) + 2 \sum_{n=1}^{\infty} (2x)^{n} \), where \( \sum_{n=1}^{\infty} (2x)^{n} = \frac{2x}{1-2x} \). Therefore, the equation becomes: \[ 2\left(A(x) - 1\right) = xA(x) + \frac{2x}{1-2x}. \] Simplifying this gives: \[ 2A(x) - 2 = xA(x) + \frac{2x}{1-2x}. \] Combining and rearranging yields: \[ (2 - x)A(x) = 2 + \frac{2x}{1-2x}. \] Thus, \[ A(x) = \frac{2 + \frac{2x}{1-2x}}{2 - x}. \] This requires some simplification: \[ A(x) = \frac{2(1 - 2x) + 2x}{(2-x)(1-2x)} = \frac{2 - 2x}{(2-x)(1-2x)}. \] Now, partial fraction decomposition can be used to express \( A(x) \) in a more manageable form. The final solutions can give you the explicit form of the sequence \( a_n \). To find \( a_n \), we'll identify the coefficient of \( x^n \) from the expression. Typically, this would involve manipulating the generating function into a recognizable form or applying techniques such as the binomial series or using known expansions for further simplification. The typical goal is to find a closed form for \( a_n \) that relates back to our original recurrence relation.