How does one deduce the approximation of $\sqrt{7}$ to be $\frac{10837}{4096}$ by taking $x = \frac{1}{64}$ in the expansion of $\sqrt{1-x}$?
How should you approach such a question? I assume the first step would be to expand $\sqrt{1-x}$ which can only be done through binomial theorem (afaik).
That gives $\sqrt{1-x} = 1 - \frac{1}{2}(x) - \frac{1}{8}(x^2)$ + ... and so on.
How do you continue? I can't seem to figure out how taking $x = \frac{1}{64}$ accomplishes anything.
HINT
Use for example
$$\sqrt{7}=\sqrt{9\cdot \frac79}=3\sqrt{\frac79}=3\sqrt{1-\frac29}$$
or according to the other hint use that
$$\sqrt{1-\frac1{64}}=\sqrt{\frac{63}{64}}=\frac38\sqrt 7$$