Answer
$\dfrac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}x^4$
$-\dfrac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2\cdot 4\cdot 6\cdot 8\cdot 10}x^5$
Work Step by Step
We are given the function:
$f(x)=\dfrac{1}{\sqrt{1+x}}=(1+x)^{-1/2}$
and the Taylor series:
$1-\dfrac{1}{2}x+\dfrac{1}{2\cdot 4}x^2-\dfrac{1\cdot 3}{2\cdot 4\cdot 6}+.....$
For $p=-\dfrac{1}{2}$ the Taylor series centered in 0 is the binomial series:
$\sum_{k=0}^{\infty} \binom{p}{k} x^k$, where $c_k=\dfrac{p(p-1)(p-2)\cdot...\cdot (p-k+1)}{k!}$
For $k=4$ we get:
$c_4=\dfrac{-\dfrac{1}{2}\left(-\dfrac{1}{2}-1\right)\left(-\dfrac{1}{2}-2\right)\left(-\dfrac{1}{2}-3\right)}{4!}=\dfrac{(-1)(-3)(-5)(-7)}{2^4\cdot 1\cdot 2\cdot 3\cdot 4}=\dfrac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}$
For $k=5$ we get:
$c_5=\dfrac{-\dfrac{1}{2}\left(-\dfrac{1}{2}-1\right)\left(-\dfrac{1}{2}-2\right)\left(-\dfrac{1}{2}-3\right)\left(-\dfrac{1}{2}-4\right)}{5!}=\dfrac{(-1)(-3)(-5)(-7)(-9)}{2^5\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5}=-\dfrac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2\cdot 4\cdot 6\cdot 8\cdot 10}$
Therefore the next two terms of the Taylor series are:
$\dfrac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}x^4$
$-\dfrac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2\cdot 4\cdot 6\cdot 8\cdot 10}x^5$
