Question

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function

Short answer

The Maclaurin series for the given function is\(\sum\limits_{n = 0}^\infty {\frac{{{2^n} + 1}}{{n!}}} {x^n}\)

Step-by-step solution

Step 1: Applied the Maclaurin series

By replacing the Maclaurin series function of given function

\(\begin{aligned}{l}f(x) = {e^x} + {e^{2x}}\\{e^x} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( x \right)}^n}}}{{n!}}} \end{aligned}\)

Replaced by in maclaurin series

Step 2: Maclaurin series function of the given function

Maclaurin series function of the given function

\(\begin{aligned}{l}{e^x} + {e^{2x}} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} + \sum\limits_{n = 0}^\infty {\frac{{{{(2x)}^n}}}{{n!}}} \\{e^x} + {e^{2x}} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} + \sum\limits_{n = 0}^\infty {{2^n}} \cdot \frac{{{x^n}}}{{n!}}\\{e^x} + {e^{2x}} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \left( {{2^n} + 1} \right)\\ = \sum\limits_{n = 0}^\infty {\frac{{{2^n} + 1}}{{n!}}} {x^n}\end{aligned}\)