Answer
$ \Sigma_{k=0}^{n} \dfrac {a^{k}x^k } {k!} $
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=e^{ax}$
Differentiate w.r.t $x$.
$f'(x)=ae^{ax}$
$f''(x)=\dfrac{a^2 e^{ax}}{2!}$
$f'''(x)=\dfrac{a^3 e^{ax}}{3!}$
Now $f(0)=1$ , $f'(0)=a$ , $f''(0)=\dfrac{a^2}{2}$, $f'''(0)=\dfrac{a^2}{3!} $ and $f''''(0)=1$
Hence, the Maclaurin Series for the function is
$\displaystyle \pi 1 +ax+ \frac {a^2} {2}x^2 + ....... \frac {a^{n}x^n } {n!} = \Sigma_{k=0}^{n} \dfrac {a^{k}x^k } {k!} $
