# Question 3b27d

Did you mean for this to be a calculus problem? Calculus can be used to prove the binomial expansion (for $| x | < 1$) (1+x)^(-1/2)=1-\frac{1}{2}x+\frac{-1/2\cdot -3/2}{2!}x^{2}+\frac{-1/2\cdot -3/2\cdot -5/2}{3!}x^3+\frac{-1/2\cdot -3/2\cdot -5/2\cdot -7/2}{4!}x^{4}+\cdots#.
${\left(1 + x\right)}^{- \frac{1}{2}} = 1 - \frac{x}{2} + \setminus \frac{3}{8} {x}^{2} - \setminus \frac{5}{16} {x}^{3} + \setminus \frac{35}{128} {x}^{4} + \setminus \cdots$ for $| x | < 1$.
This means $\setminus \frac{1}{\setminus \sqrt{.99}} = {\left(1 + \left(- .01\right)\right)}^{- \frac{1}{2}} = 1 + \frac{.01}{2} + \setminus \frac{3}{8} {\left(.01\right)}^{2} + \setminus \frac{5}{16} {\left(.01\right)}^{3} + \setminus \frac{35}{128} {\left(.01\right)}^{4} + \setminus \cdots$
$\setminus \approx 1 + 0.005 + 0.0000375 + 0.0000003125 + 0.00000000273438 \setminus \approx 1.00504$