The problem defines a binary operation $*$ on the set of real numbers $R$ such that $a * b = a + b + a^2b$. We are asked to evaluate $\frac{1}{4} * \frac{1}{2}$.

1. Problem Description

The problem defines a binary operation $*$ on the set of real numbers $R$ such that $a * b = a + b + a^2b$. We are asked to evaluate $\frac{1}{4} * \frac{1}{2}$.

2. Solution Steps

We are given that $a * b = a + b + a^2b$. We want to find $\frac{1}{4} * \frac{1}{2}$, so we substitute $a = \frac{1}{4}$ and $b = \frac{1}{2}$ into the expression for $a * b$.

$\frac{1}{4} * \frac{1}{2} = \frac{1}{4} + \frac{1}{2} + (\frac{1}{4})^2 (\frac{1}{2})$

Now we simplify:

$\frac{1}{4} * \frac{1}{2} = \frac{1}{4} + \frac{1}{2} + (\frac{1}{16})(\frac{1}{2})$

$\frac{1}{4} * \frac{1}{2} = \frac{1}{4} + \frac{1}{2} + \frac{1}{32}$

To add the fractions, we need a common denominator, which is

3

2. $\frac{1}{4} = \frac{8}{32}$

$\frac{1}{2} = \frac{16}{32}$

So, we have:

$\frac{1}{4} * \frac{1}{2} = \frac{8}{32} + \frac{16}{32} + \frac{1}{32}$

$\frac{1}{4} * \frac{1}{2} = \frac{8+16+1}{32}$

$\frac{1}{4} * \frac{1}{2} = \frac{25}{32}$

3. Final Answer

$\frac{25}{32}$