Answer
$\Sigma_{k=0}^{n} \dfrac {x^{2k+1}}{(2k+1)!}$
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!} + ..\\
$
$f(x)=\sin h x=\dfrac{e^x-e^{-x}}{2}$
Differentiate w.r.t x
$f'(x)=\dfrac{e^x+e^{-x}}{2}$
$f''(x)=\dfrac{e^x-e^{-x}}{2}$
$f'''(x)=\dfrac{e^x+e^{-x}}{2}$
$f''''(x)=\dfrac{e^x-e^{-x}}{2}$
Now $f(0)=0$, $f'(0)=1$, $f''(0)=0$, $f'''(0)=1$ and $f''''(0)=0$
Hence the Maclaurin Series for the function is
$\displaystyle x+ \frac {x^3} {6} + .......+ \frac {x^{2n+1}}{(2n+1)!} = \Sigma_{k=0}^{n} \frac {x^{2k+1}}{(2k+1)!}$
