1. **State the problem:** We want to write the Taylor series of the function $f(x) = e^{2x}$ about the point $x_0 = \frac{1}{2}$ using sigma notation. 2. **Recall the Taylor series formula:** $$ f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k $$ where $f^{(k)}(x_0)$ is the $k$-th derivative of $f$ evaluated at $x_0$. 3. **Find the derivatives:** Since $f(x) = e^{2x}$, each derivative is: $$ f^{(k)}(x) = 2^k e^{2x} $$ 4. **Evaluate derivatives at $x_0 = \frac{1}{2}$:** $$ f^{(k)}\left(\frac{1}{2}\right) = 2^k e^{2 \cdot \frac{1}{2}} = 2^k e^{1} = 2^k e $$ 5. **Substitute into the Taylor series formula:** $$ f(x) = \sum_{k=0}^\infty \frac{2^k e}{k!} (x - \frac{1}{2})^k $$ 6. **Final answer:** $$ e^{2x} = e \sum_{k=0}^\infty \frac{2^k}{k!} (x - \frac{1}{2})^k $$ This is the Taylor series expansion of $e^{2x}$ about $x_0 = \frac{1}{2}$ in sigma notation.