Given that $|x|\gt2$ find the first four terms in the series expansion of $\left(1-\frac2x\right)^{\frac12}$ in descending powers of $x$. By taking $x$ = 200 use the series to find a value of $\sqrt{99}$, giving your answer to 7dp. Use your series to find $\sqrt{101}$ to the same degree of accuracy.
I have obtained the series and made $x$ = 200 $$1-x^{-1}-\frac{x^{-2}}{2}-\frac{x^{-3}}{2}-...$$ $$1-0.005-0.000012-0.0000000625$$ $$=0.9949879375$$
What is the intuition to multiply this result by 10 to get the answer of $\sqrt{99}=9.9498792...$
and how do you apply it to $\left(1-\frac2x\right)^{\frac12}$ to get $\sqrt{101}$?
For the first part, if you put $x=200$, your left hand side becomes $$\sqrt{1-\frac{2}{200}}=\sqrt{\frac{99}{100}}=\frac{\sqrt{99}}{10}$$ For the second part, you want your $-2/x$ to equal $+1$. What must $x$ therefore be?